### vovuh's blog

By vovuh, history, 4 years ago,

All ideas belong to MikeMirzayanov.

1399A - Remove Smallest

Tutorial
Solution

Tutorial
Solution

1399C - Boats Competition

Tutorial
Solution

1399D - Binary String To Subsequences

Tutorial
Solution

1399E1 - Weights Division (easy version)

Tutorial
Solution

1399E2 - Weights Division (hard version)

Tutorial
Solution

1399F - Yet Another Segments Subset

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Solution
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 » 4 years ago, # |   +12 Why dsu didn't work in problem F.89050221i am finding all invalid pair and add a edge between them ...then i am finding the maximum independent set;
•  » » 4 years ago, # ^ |   +30 Actually, maximum independent set is pretty hard problem to solve and can be done in $O(n^3)$ using Kuhn's algorithm on bipartite graphs. The graph you build isn't even bipartite (you can easily obtain odd length cycle with segments $[1; 5]$, $[4; 8]$ and $[3; 6]$, they all have pairwise intersections). I don't know a good solution for this problem for non-bipartite graphs, just some randomized approaches which don't work well.
•  » » » 4 years ago, # ^ |   0 thanks.
•  » » » 4 years ago, # ^ |   0 Would this work for F?.First, co-ordinate compress the segments. After that, sort these by ending indices. Maintain a dp array where dp[i] = size of optimal subset where all segments have ending indices <= i. Consider the Kth segment to be the last segment included in our optimal subset and it's starting and ending indices to be L and R respectively. Loop through the points from R->0 on the co-ordinate axis, then, dp[R] = max(dp[R], number of segments inside range [L,R] + dp[L-1]);The maximum of dp[i] would be our answer. I think this is the same idea as the editorial, couldn't understand that one exactly, so came up with this.Note: The kth segment is considered to be inside [L,R].
•  » » » » 4 years ago, # ^ |   0 Yes it works. Also, you don't need to coordinate compress. Here's the submission: 89075517
•  » » » » » 4 years ago, # ^ |   0 Isn't the solution you wrote O(n) complexity? If I am mistaken please correct me
•  » » » » » » 4 years ago, # ^ |   0 In the solve function I'm also doing binary search, which adds a log factor.
•  » » » 4 years ago, # ^ |   -43 $O(n^3)$?A set is independent iff its complement is a vertex cover, so this is the same as computing minimum vertex cover. And by Konig theorem in bipartite graph minimum vertex cover equals maximum matching which can be computed faster than $O(n^3)$. You can of course construct a weighted bipartite graph with cost 1 if there was an edge in original graph or 0 otherwise and run maximum weighted matching (Kuhn a.k.a. Hungarian algorithm) but that's just insane.
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +13 I don't get your point.Yes, thanks, I know about Konig's theorem and know that vertex cover and independent set (and maximum matching) can be converted to each other.I didn't talk about Hungarian algorithm, I talked about dfs-like Kuhn algorithm for finding unweighted maximum matching. It's time complexity is $O(n^3)$ also. In practice, it's way faster, yes, but the complexity is still $O(n^3)$ (or, to be precise, $O(nm)$).And you can construct clique in this problem pretty easy, so Dinic or Hopkroft-Karp would work with the same time complexity as this unweighted Kuhn matching.Edit: Sorry, Dinic and Hopkroft-Karp will work a bit better ($O(m \sqrt{n})$ instead of $O(nm)$).
•  » » » » » 4 years ago, # ^ |   -8 Google-search phrase "Khun algorithm" and you'll find out why I got confused by your previous comment
•  » » » » » » 4 years ago, # ^ |   0 Yes, I googled and this is why I tried to explain to you what am I talking about. I'm really surprised I didn't find any explanation of this algorithm in English (didn't go deep into search but anyway). I always thought this is very famous approach but now I realized all articles and lessons I heard about it were from Russian teachers.
•  » » » » » » » 4 years ago, # ^ |   -15 Of course this approach is well known. It's simply that the algorithm you are talking about is not called "Khun algorithm", but "Hopcroft-Karp" or "turbo matching", Moreover simce the actual "Khun algorithm" aka Hungarian does run in $O(n^3)$ whilst Hopcroft Karp doesn't I did assume you actually mean Khun algorithm and got really confused on rest of your comment
•  » » 4 years ago, # ^ |   -47 Video Editorial of Problem D: Binary String to Subsequences
•  » » 4 years ago, # ^ |   +9 I had the same approach. At least i'm glad to know that i wasn't the only one. Thanks vovuh for the explanation :)
•  » » 4 years ago, # ^ |   +9 I tried the same and then I learned that finding maximum independent set is not as easy as I thought
•  » » 4 years ago, # ^ |   0 Maybe because it doesn't have transivity.For example,Segment A=[2,3],B=[3,4],C=[4,5], A-B ans B-C are invalid,but A-C is valid.So you can't map in this way.
 » 4 years ago, # | ← Rev. 12 →   0 Video Explanations(hindi) - Question D- https://youtu.be/BRrPfZQ3ScU - Question C- https://youtu.be/txEX3EwWaSw - Question B- https://youtu.be/NxfA91zSNlI - Question A- https://youtu.be/tSboQu6Vu_k
•  » » 4 years ago, # ^ |   +72 Video explanations to everything else. (And also A-C lol)
•  » » » 4 years ago, # ^ | ← Rev. 4 →   +6 Your videos are really cool, have been following & learning from you
•  » » » » 4 years ago, # ^ |   +45 But you didn't learn the main thing, he started YT when he was already very good at cp.
•  » » » » » 4 years ago, # ^ | ← Rev. 3 →   +56 I made a YT while i am becoming good at CP :-PI agree i am still learning, but i only make videos on the questions i am able to solve and 100% sure about. Its not like i am pretending that i have full knowledge and am god at cp. 1 month ago i was not able to solve 1 question.But after this YT, there is a source of motivation for me to learn, and now am able to solve more so i am happy.:-)
•  » » » 4 years ago, # ^ |   -22 Video Editorial of Problem D: Binary String to Subsequences
•  » » » » 4 years ago, # ^ |   +3 I always see that this guy gets a downvote whenever he posts the link for the tutorial. why so ??
•  » » » » » 4 years ago, # ^ |   +8 i was thinking the same, his explanation for D was nice, and for the first time i also posted today, got -14 votes, but its balanced now to 0. IDK why.
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   -7 .
•  » » » » » » » 4 years ago, # ^ | ← Rev. 2 →   +11 I completely understand this, but i make videos on what i am able to solve on my own, and those which i cant i don't even try to make as i dont want to share incorrect knowledge.But why this dude you talked about, i was curious about him
•  » » » » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 .
•  » » » » » » » » 4 years ago, # ^ |   +5 TBH, I really like your idea of uploading videos irrespective of your rating. You want to make videos and hence you are making them. You aren't forcing anyone to watch your videos. If I had been in your shoes, I would never have the courage to start a channel and hence I'm really proud of the fact that you are uploading videos.As per why my comment is getting so many downvotes, I myself have no idea. I'm getting good enough response on YouTube. Maybe someone who downvoted my comment can tell me why they did it and I can keep that in mind from next time?
•  » » » » » » » » » 4 years ago, # ^ |   0 I don't like the impression of making YT videos and commenting here twice (or maybe even once). I hate monetization of everything (specially educational) and I do believe better way to learn is try yourselves and if you have exhausted all thoughts read editorial (because its not spoon fed here).On top of that commenting it in comments section of an official editorial irritates me.I know I don't have completely sound arguments so I'd normally just down vote and move.
•  » » » » » » » » » 3 years ago, # ^ |   +3 brother if you have guts then please can you explain me solution's like sKSama and namanbansal013 does. After every contest which I give I wait for their videos. If You have guts than please we'll welcome you!
•  » » » » » » » » » 3 years ago, # ^ |   0 I didn't mean to offend, Its perfectly fine to do so. Its just the way I see things.To be specific I have issues with this kind of promotion (which again is fine but I don't like it).
•  » » » » » » » » » 3 years ago, # ^ |   0 it's fine if you don't like it but don't demotivate them at least they are doing something if you don't like then ignore but don't downvote them their effort's are on another level!
•  » » » » » » » » » 3 years ago, # ^ |   0 If I had disliked his content on youtube then what you say would be appropriate. But he had promoted his content here (twice) which I didn't like and only thing I can do to show that is downvote.
•  » » » » » » » 4 years ago, # ^ |   -10 I completely dont agree with the claim that a division 2 user shouldn't post explanations. Before you downvote, let me explain why. Firstly, some people are doing other websites than codeforces and have good ratings there. There are different systems, some speed based like cf, some more OI style.Secondly, better rating does not imply better explanation hust as worse rating does not imply worse explanation. Some people are very good at explaining solutions to things they solve and even if they are not high rated it's a good source.Third, I believe that a good editorial explains how one would approach a problem. Some things that are considered trivial by div1 users are not considered trivial by a div2 user, and perhaps an explanation from a div2 user would be more practical for a div2 user.Fourth, some people use alts here or simply were high rating then dropped. Fifth, about the "waste of time", as Feymann once said, the best way to check you understand something is to make someone else understand it aswell, and it also helps so why not?Sixth, some people are just slow but are able to solve high rated problems, you shouldnt be ratist in general, not just in this case.Seventh, as a specialist you dont have the right to make ratist arguments :)If you have balls, I offer you a 2300 codeforces duel, now stop demotivating the community and start solving more problems :)
•  » » » » » » » » 4 years ago, # ^ | ← Rev. 3 →   0 If you downvote, the least you can do is reply with a valid counter-argument, otherwise you are just ratist :)Edit: So apparently, atleast 7 people saw rating and downvoted without a counter-argument. Either write one or duel me on codeforces. Some IOI medalists dont even do codeforces. You gave zero thought, saw blue (was CM before btw) and immediately downvoted. Very mature.
 » 4 years ago, # | ← Rev. 2 →   0 https://mirror.codeforces.com/contest/1399/submission/89052941Q-3 Can someone tell me what's wrong with this solution ?
•  » » 4 years ago, # ^ |   0 In the first for loop you are doing sum[a[i]+a[j]]++ which is wrong because lets say there are 3 people with weights {1,3,3} In the first iteration , you are using the zeroeth person thrice (when (i,j) is (0,0) (0,1) (0,2)). In the first iteration itself sum[4] will be 2 which is not possible since if the zeroeth person made a team with the first person for sum=4, he cannot make another team with the second person. As a person cannot be in 2 teams, this approach fails.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 I did that in the last attempt sir, previously i used for(j=i+1;j
•  » » » » 4 years ago, # ^ |   0 Yes, you have to select the weight with the maximum frequency. But the way you are creating the sum array is wrong because that loop is running n^2 times and creating almost n^2 pairs. The problem is that when you hash them to the sum variable, you are using one person more than once (in different pars), which leads to a wrong sum array.
•  » » » » » 4 years ago, # ^ |   0 what if I change the second array to for (j=i+1;j
•  » » » » » » 4 years ago, # ^ |   0 Nope. AFAIK you cannot create a hash sum array for this question. You have to create a freq array of numbers and then run a loop from (2 to 2*n) and then search on the freq array for the sum, as given in the editorial.
 » 4 years ago, # | ← Rev. 2 →   -6 I want to know the approach of E1 if it is slightly changed so that we have to find the minimum number of operations so that distance between the root and every leaf is less than equal to S.I have an approach in mind to this problem but I'm not sure if it would work. Can anyone please help me out. UPD : vovuh can u please provide your crucial insights to this.
 » 4 years ago, # |   0 Please tell me what's wrong in my E1 solution
•  » » 4 years ago, # ^ |   +11 I also did like this only. The only mistake I spotted in your code is that in your priority queue, you're arranging the edges by (weight * nx) [nx = no of leaves for this edge contributes to the sum]. Instead, it should be the reduction we can get if we make our operation in it, i.e. (weight * nx — (weight / 2) * nx). Example to prove the contradiction, let's say weight1 = 7 and nx1 = 3, and weight2 = 3 and nx2 = 7. If we do the operation on 1st one, 7*3 = 21 will reduce to 3*3 = 9, hence the deduction is 12. On the other hand if we do on 2nd one, 3*7 = 21 will reduce to 1*7 = 7, hence the deduction is 14. But in your code, 1st one will be operated. Hope you understood. My submission — 89053607
•  » » » 4 years ago, # ^ |   0 Thanks so much!!!
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 Wait, if I arrange like this {weight * nx, nx,weight} why it's not correct again? then {21, 7, 3} has a higher priority than {21, 3, 7}. tbh, i didnt really consider this case, but i think my code does not fail here.
•  » » » 4 years ago, # ^ |   0 I would like to know, why ordering by weight*nx does not work? Thanks in advance.
•  » » » » 4 years ago, # ^ |   +1 See I've explained a contradictory case [{21, 7, 3} and {21, 3, 7}] in my comment above.
•  » » » » » 4 years ago, # ^ |   0 but if the second element in this triple is nx, 3, 7 and 7, 3 is not a contradiction. See my comment above.
•  » » » » » » 4 years ago, # ^ |   +3 No, it won't work like that. You've to keep the parameter which actually matters after doing the operation, i.e. the deduction it is making. I can give contradictory cases to fail this too. For example, weight1 = 8 and nx1 = 9, & weight2 = 9 and nx2 = 8. You say order by nx, then 1st one will be chosen, hence 4*9 = 36, the deduction will be 36, whereas if you take 2nd one, 4*8 = 32, the deduction will be 40.
•  » » » » » » » 4 years ago, # ^ |   0 Finally got it, Thanks.
 » 4 years ago, # | ← Rev. 2 →   +59 For F, we also have a more memory efficient solution with the same time complexity. First, let's sort all the segments in order of increasing segment length. Now, let $dp[i]$ represent the maximum number of segments we can take (including the current one), such that all of them are completely covered by the $i$th segment and they abide by the given constraints. To compute each $dp[i]$, we only need to consider segments that are completely covered by it (by definition). Notice that these segments must have length strictly less that the length of the $i$th segment, so we already know all of their $dp$ values. Therefore, to get the value of $dp[i]$, we will run normal segment dp on the set of covered intervals (finding the largest set of intervals such that no two overlap, where the size of an interval is its $dp$ value instead of just $1$). For each segment, computing its $dp$ value takes $\mathcal{O}(N)$, so our final time complexity is $\mathcal{O}(N^2)$. To compute the answer more easily, we can insert an imaginary $N+1$th segment with endpoints $[-10^9, 10^9]$, with the final answer being $dp[N+1] - 1$. The advantage of this solution is that we only need to consider the $\mathcal{O}(N)$ segments that are covered by segment $i$ when computing $dp[i]$, so our memory complexity is only $\mathcal{O}(N)$, which is easily within $256\textrm{MB}$.
•  » » 4 years ago, # ^ |   0 Tried implementing a similar idea to achieve $O(N)$ memory complexity but my solution is TLEing at the last test case.Here's the idea: Sort according to the end points of the segments and let $dp[i]$ = the maximum number of segments you can pick considering only the first $i$ (in the sorted list).The transition involves trying to pick the $i^{th}$ segment, which requires running exactly the same dp on all subsets of the $i^{th}$ segment. Then compare this to not picking the $i^{th}$ segment, which corresponds to just considering the first $i-1$ segments.I'm not sure of the time complexity, but I believe it should be $O(N^2)$. Would appreciate if somebody double checks this/gives ideas for optimization.Here's the submission: 89072684 Commented Code:#include #define rep(i,n) for(int i=0;i<(n);i++) #define repl(i,a,b) for(int i=(a);i<(b);i++) struct range { int l, r, ind; }; int rangedp[3005]; vector subs[3005]; int solve(vector lr, int ix=3001) { if(lr.size()<=1) return lr.size(); if(rangedp[ix]!=-1) return rangedp[ix]; int n = lr.size(); int dp[n+1]; dp[0]=0; repl(i, 1, n+1) { dp[i]=1; // picked this one // solve for elements inside current segment dp[i]+=solve(subs[lr[i-1].ind], lr[i-1].ind); // binary search last index that doesn't // intersect with current segment int l=0, r=i-1, j=i-1; while(l<=r) { int mid=(l+r)>>1; if(lr[mid].r>=lr[i-1].l) { j=mid; r=mid-1; } else l=mid+1; } // add the number of segments you can get // by including the j segments that don't // intersect with the current one dp[i]+=dp[j]; // is it optimal to include current segment // or should we skip it? dp[i]=max(dp[i-1], dp[i]); } return rangedp[ix]=dp[n]; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); int t; cin >> t; while(t--) { int n; cin >> n; vector lr; rep(i, n) { range val; cin >> val.l >> val.r; val.ind=i; lr.push_back(val); } // sort according to end point, break tie so that // larger segments appear later sort(all(lr), [](range a, range b) { if(a.r!=b.r) return a.r < b.r; else return b.l < a.l; }); // precompute subsets of each segment rep(i, n) subs[i].clear(); repl(i, 1, n) { int j=i-1; while(j>=0 && lr[j].r>=lr[i].l) { if(lr[j].l>=lr[i].l) subs[lr[i].ind].push_back(lr[j]); j--; } reverse(all(subs[lr[i].ind])); } memset(rangedp, -1, sizeof(rangedp)); cout << solve(lr) << endl; } return 0; } 
•  » » » 4 years ago, # ^ |   +3 You're copying vector everytime you call it, pass it by a reference.
•  » » » » 4 years ago, # ^ |   0 That fixed it, thanks!
•  » » 4 years ago, # ^ |   0 Can you explain how to find thisfinding the largest set of intervals such that no two overlap
•  » » » 4 years ago, # ^ |   0 Please see here.
•  » » » 4 years ago, # ^ |   +3 We can sort the segments by their endpoints and run a dp, where dp[i]= Maximum value such that ith segment is included or not included in the optimal answer.For the transitions, we can use binary search to find the last segment which doesn't overlap with ith segment, let's call it prev.So, dp[i]=max(dp[i-1],value[i]+dp[prev]). My code in Java: 89147950
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 How do you calculate value[i]? Edit:- Never mind, got it. Using a different sort for finding value[i].
 » 4 years ago, # | ← Rev. 2 →   -8 My simple solution for C:Let's create a function to get the maximum number of teams with weight m. In the function we iterate over all pairs of numbers and if their sum is equal to m and the numbers were not used before, mark the numbers and update the answer. Next, let's go over all possible weights. Due to small restrictions, such values are few. The answer to the problem will be the maximum of all such values88999789
 » 4 years ago, # | ← Rev. 2 →   +60 Initially I misread problem E. I thought of a different version for the problem and wasted 30 min of my time. The problem I thought of was to print the minimum number of operations so that max(w(root,v))<=S where v belongs to the set of leaves of the tree. Is this problem solvable? If yes, then can somebody suggest a solution for this version of the problem?
•  » » 4 years ago, # ^ |   +8 Haha I actually made the exact same mistake (then they came out with the clarification and I felt like such a fool... not the first time I've misread a problem). Didn't make much progress on that though.
 » 4 years ago, # |   0 I attempted problem C while taking the input in a vector (https://mirror.codeforces.com/contest/1399/submission/89066257) and in an array (https://mirror.codeforces.com/contest/1399/submission/89066399). The vector approach gives me a TLE but the array approach passes all test cases within time limit, I am allocating the memory at the start in a vector and not even using the push_back() function, is there any particular reason for this?'
•  » » 4 years ago, # ^ |   +1 You don't handle n=1 correctly and read outside of array bounds leading to strange values of s1 and s2. Because the array and vector have different sizes on the stack the memory read by a[1] and a[n-2] is different between your solutions so you end up with different behaviour.
 » 4 years ago, # |   0 My answer in the checker is correct, butI am getting the following comment from the checker: wrong output format Expected int32, but "7882141163342456" found (test case 1)https://mirror.codeforces.com/contest/1399/submission/89012361
•  » » 4 years ago, # ^ |   +1 your ans is not correct
•  » » » 4 years ago, # ^ |   0 See the jury's answer and contestant's answer, they both are same. And in the correct submission after that, I did not change my logic, just changed long long int to int
•  » » » » 4 years ago, # ^ |   +25 huh, 1 2 3 1 1 4 5 6 7 8 1 1 9 10 11 1 1 1 1 1 1 12 1 1 13 14 15 16 1 2 1 2 17 7882141163342456 7882141163342456... and 1 1 2 3 3 3 4 5 6 7 8 8 8 9 10 11 11 11 11 11 11 11 12 12 12 13 14 15 16 16 15 15 16 17 17 17 18 19 19... are exactly the same.
•  » » » » » 4 years ago, # ^ |   0 Oh sorry mybad, I misread something. I was checking only the 100002 part. Sorry for the hassle
 » 4 years ago, # |   +2 What a fast editorial! Really helpful; Thank you!
 » 4 years ago, # |   +3 Why is the complexity of a heap query in E1 log(nlogw) and not just log(n)? The number of items in the heap is constantly n so why is it not log(n)?
 » 4 years ago, # |   -7 Does anyone have any other approach for E2
•  » » 4 years ago, # ^ |   +1 Before all, we calculate the number of leaves in every subtree using DFS. The contribute of the lowest edge between the subtree's root and the root is the number of leaves in the subtree multiplied by the edge's weight. Recall the contribute of one move is the diff of its edge's contribute after move. For example, for an edge of weight 5 whose subtree has 3 leaves, its contribute is $5*3=15$, and the contribute of its first move is $5/2*3-5*3=-4$.We use a multi-set/priority_queue or other data-structure containing all available moves sorted by its contribute in non-decreasing order. When we do a move, we push the next move in data-structure (weight divided by 2). If two moves' contribute equal, which first? We first take the one of which next move has lower contribute. If the next move equal, we check the next of next, and so on. But this take too long time($\log W$), we have better method. If two moves' contribute equal, either their's weight - weight/2 equal and the leaves number of their's edge equal or any one's weight-weight/2 equals the other's leaves number. First case, one has large weight will always win. Second case, one has little weight will always win. You can prove it easily.There're exactly 2 cases of the answer, odd or even. If the answer is even, we can think two one-cost-move as one two-cost-move, and do this process greedily. If the answer is odd, we can take a one-cost-move that has lowest contribute. This is my submission: 89086737.I'm not sure whether it correct, although it get Accept before system test.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 Nice approach I really liked it thanks.
•  » » » » 4 years ago, # ^ |   0 And I think there's another approach. Do the process in all one-cost-move, and in all two-cost-move separately. We can get the minimal contribute (contribute is non-positive) of every possible cost using only one-cost-move, and the minimal contribute of every possible cost using only two-cost-move. And enumerate the number of one-cost-move, find minimal number of two-cost-move and update answer.
•  » » » » » 4 years ago, # ^ |   0 So as in E1 we calculate cost 1 edges and 2 edges using same approach and then find the answers separately and then try minimising 1 cost answer by keep on replacing 3 of the 1 cost moves with 2 cost moves?
•  » » » » » » 4 years ago, # ^ |   0 sorry, I haven't tried E1...
•  » » » » » » » 4 years ago, # ^ |   0 E1 is just E2 with cost of every edge is 1
•  » » 4 years ago, # ^ |   +20 Iam sorry if someone felt offended by my comment. But I didn't have such intentions. I don't get why people down vote others comments. I am not spreading rumours or criticizing someone or putting something which hurts someone. I am really trying my best to grow and help others to grow along with me. Do you all know how ugly it would be if anyone including my recruiters who open my profile and see a negative contribution. Guys we are members of the same community please try to help others not critisize them. If anyone felt offended by this message iam sorry again.
•  » » 4 years ago, # ^ |   0 Can you please share your approach for E1
•  » » » 4 years ago, # ^ | ← Rev. 3 →   0 At first we have to infer the following from the question. if we perform an operation on an edge it will effect all the paths in which it is there (cost will be (E/2) *N) where E is the edge weight and N is the number of paths in which the edge is present we have to minimise the no of moves we make implies we have to always pick the edge which has maximum cost. So once he have the above intuition we can decide to calculate no of paths a particular edge is in using simple dfs traversal. Then we have to choose a data structure which has the following properties maintains the sorted order less time for insertion and removal So from the above properties we use Max-heap (priority queue) in c++.We then insert every edge into priority queue First parameter being the cost to operate on edge i.e,(E/2) *N where E is the edge weight and N is the number of paths in which the edge is present Second parameter being cost Third parameter no of paths.We keep on doing the following operations until effective cost is less than s(given in the question). remove maximum in the heap perform Operation on it insert updated edge here is the link to my submission
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 hey thank you for replying : https://mirror.codeforces.com/contest/1399/submission/89208650 (I am getting an TLE while using set() to accesing the diff and AC when using getdiff(id) ) Can you please explain Why AC:https://mirror.codeforces.com/contest/1399/submission/89208760
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Don't know sorry
 » 4 years ago, # | ← Rev. 3 →   +11 My video solutions to all problems + screencast (where I placed 17th? place). Sorry for uploading it that late (currently slow internet connection)Enjoy watching.https://youtu.be/yWD-Hcl_0SkUPD:After a system testing, I'm 15th
•  » » 4 years ago, # ^ |   0 thank you so much
 » 4 years ago, # |   0 Why editorial before closing hacking phase?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +1 hacking doesn't effect rating in div 3. so people can learn to hack here.
•  » » » 4 years ago, # ^ |   +3 Well not exactly. If you successfully hack submissions of those better in the ranking, then you get a lower rank and subsequent increase in rating. But yes, there is little difference (unless there is massive hacking for a particular question), because unlike Div 2, we don't get points for successful hacking...
•  » » » » 4 years ago, # ^ |   0 thanks for the info
•  » » » » 4 years ago, # ^ |   0 Doesn't matter if someone with a faulty solution is not hacked because it will eventually fail the system testing.
 » 4 years ago, # |   0 Can someone help me with Problem D — my solution
•  » » 4 years ago, # ^ |   0 Where do you want help here, it's completely incorrect.
 » 4 years ago, # |   0 #include using namespace std; int main() { cout<<1000<
•  » » 4 years ago, # ^ |   -9 You are probably using this code on some online compiler, copying the output from there and then pasting it as manual input. But online compilers doesn't really give you complete output for such big codes, they probably stop at around 500th test case among your 1000 cases (you can check that by writing cout<
•  » » » 4 years ago, # ^ |   0 i am using codeforces generator.
 » 4 years ago, # | ← Rev. 3 →   0 https://mirror.codeforces.com/contest/1399/submission/89071282 Problem E1 Giving TLE in test case 2 approach same as that of editorial can anyone explain .why?
 » 4 years ago, # |   +1 Submission : 89071344 Image : (https://ibb.co/x1sGJ18)The submission, I have made for 1399E1 - Weights Division (easy version), gets AC. However, as I run it on my Sublime IDE, I get the following (error? maybe, I'm not sure exactly) : ErrorTime taken : 0.001311s free(): invalid pointer timeout: the monitored command dumped core bash: line 1: 38557 Aborted timeout 2.0s '/home/imperial_lord/Sublime/main' < input.txt > output.txt [Finished in 1.8s with exit code 134] [cmd: ['bash', '-c', "g++ -std=c++17 '/home/imperial_lord/Sublime/main.cpp' -o '/home/imperial_lord/Sublime/main' && timeout 2.0s '/home/imperial_lord/Sublime/main' output.txt"]] [dir: /home/imperial_lord/Sublime] [path: /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games:/snap/bin] (You can also see the image link).I had a similar error, during the contest and after I decided to upsolve. I feared a WA, so I didn't try submitting, but could not debug. Now I have made my solution similar to the author's, but the issue still persists. Could someone take a look? I have not cleared any vectors, neither is my pointer invalid for the set, so I'm not sure, what free(): invalid pointer means. P.S.It was a nice problem-set, really enjoyed it!
•  » » 4 years ago, # ^ |   +1 Try compiling with the flags -g -D_GLIBCXX_DEBUG. It will catch most common errors in STL containers and provide some helpful information about them.In this case, you are trying to dereference an iterator after you just erased its contents from a set. The error produced by the debugger is "Error: attempt to dereference a singular iterator.".
•  » » » 4 years ago, # ^ |   +1 Thanks a lot for the reply. I'll definitely keep this flag in mind. However, that was not what was causing the error for me. Why?I simply commented out the while loop part and the printing of the answer part. The error was still there.But, after doing some work, I finally found it. It was in the DFS function I had written. With Errorvoid dfs(ll v, ll p=-1) { if(adj[v].size()==1) cnt[p]=1; ...  Without Errorvoid dfs(ll v, ll p=-1) { if(adj[v].size()==1 && p!=-1) cnt[p]=1; ... So cnt[-1] was the issue. I'm not sure if this should cause an issue in some testcase, because the editorial has same code.
 » 4 years ago, # |   0 Can anyone please help me in explaining,what is the thought process behind D like how are you able to arrive at the solution?
•  » » 4 years ago, # ^ |   0 Well, I haven't read the editorial yet, because I got an AC during the contest itself, but here is how I approached the problem. So, we got a string of 1s and 0s and the length is n. The task is to make partitions, i.e. divide it into subsequences which are either of the form 10101.. or 01010...So let me assign group 1 to my first character. And simultaneously, I'll try to decide what is the next character my group 1 needs. ObviouslyIf s[0]='1' this means my group 1 will need a 0, otherwise it'll need a 1.Now here comes the second element. I just need to check, whether it'll fulfill the hunger of group 1 (that is it gives the required element), or whether it'll just make another group 2 of it's own and wait for a suitable match. This is the way, we need to do for all the rest of the positions.Now to solve it effectively, I just need to maintain the group numbers for those waiting for 0s and 1s, and so I'll need a map and I'll make a map of sets, because of it's O(n) time insert and delete (I'll need these operations, because the groups will keep swapping there needs between 0s and 1s).My Submission : 89003229
 » 4 years ago, # |   0 Congratulations on great editorial!
 » 4 years ago, # | ← Rev. 2 →   0 In problem E1 I got tle on test case 2 when used set and after replacing set by priority queue it got AC in 0.2 sec.Can somebody explain it clearly how too much time differenceUpd: faker_faker thanks got it!!
•  » » 4 years ago, # ^ |   0 set doesn't have duplicates, so you go into infinite loop.AC on making it multiset: 89075373
 » 4 years ago, # | ← Rev. 2 →   +12 Greedy solution to E2 using two priority queues (PQ1 for cost 1 edges, PQ2 for cost 2 edges). My while loop logic (sum > S) was (select 1 of the following):1) If decreasing the best cost-1 edge is sufficient, do so and end immediately.2) If decreasing the best cost-2 edge is sufficient, do so and end immediately. Note doing this check after check (1) is safe, since if (1) failed, we need to pay at least 2 in any case.3) Easy case: if either of the priority queues is empty, you have to choose the one that has edges.4) Determine the reduction in sum (call RED1) if you choose edges from PQ1 twice. (Edge case: remember that the second edge chosen might be the same as the first edge, after reducing by half). Determine the reduction in sum (call RED2) if you choose one edge from PQ2. If RED1 >= RED2, then perform the operation on the best edge in PQ1. Else perform the best edge in PQ2.
•  » » 4 years ago, # ^ |   0 In (4) is it not sufficient to check if RED2/2 >=RED1(here red1 is choosing from pq1 once) and if it is true perform operation on pq2 else on pq1 ? can you give a counter example of it?
•  » » » 4 years ago, # ^ |   0 The proof for my original logic seemed more straightforward whereas I couldn't quickly reason why your alternative (which I also considered) would work. And indeed, I get WA when trying the alternative.Unfortunately, I can't seem to think of a (small) counterexample.
•  » » » » 4 years ago, # ^ |   +5 It turns out that this alternative is also correct. I tried this approach and got WA on test 21(like you). I also managed to prove it but still something was wrong. Finally, after a few hours of analysing this problem I found out that I(and you) was missing a one little case when the last edge you are decreasing is a cost-2 edge and you can undo the last cost-1 edge-decreasing. Hope that helps :D
 » 4 years ago, # |   0 On the standings page, what is the difference between a “+” and a “+1” for a problem?
 » 4 years ago, # |   +1 Does anyone have a working python solution for E1?
 » 4 years ago, # | ← Rev. 2 →   0 Why 89023478 got a TLE?(Solution to D) Can anyone tell please?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +5 You are erasing the first element of vector. vector.erase() has a time complexity equal to the length of the vector because when you delete the first element you have to shift the entire vector to the left one position. As you're calling erase() multiple times inside a loop, when the vector length becomes large, you get TLE. Instead of erasing the first element, you can erase the last element using pop_back() which has a constant time complexity because the order in which you insert '0' and '1' in the existing subsequences doesn't matter (as long as you're maintaining minimum number of subsequences).
•  » » » 4 years ago, # ^ |   0 Thank you bro! I couldn't find why! Thanks a lot!
 » 4 years ago, # |   0 In problem D,each character of the string belongs to exactly one subsequence. In the second line print n integers a1,a2,…,an (1≤ai≤k), where ai is the number of subsequence the i-th character of s belongs to. Ex: 4 0011 Ans: 2 1 2 2 1 I don't understand where 1 2 2 1 (second line print) come from. You divide 0011 to 01 and 01 ( 2 subsequence ) and each character should belongs to 1 subsequence. So should it print 1 1 1 1 instead of 1 2 2 1? Am I misunderstood something?
•  » » 4 years ago, # ^ |   +1 .....where ai is the (index) number of (the) subsequence, the i-th character of s belongs to.Even if it wasn't clear to you, it was kinda obvious :)
•  » » 4 years ago, # ^ |   +1 Actually here, all the characters(0 or 1) are belong to exactly one sub-sequence. So, from your point of view the result should always be 1 1 1 1... right? But no, actually here it's meant that if there are n sub-sequences, number them from 1 to n. then print which character belongs to which sub-sequence. For example: 4 0011 has 2 valid sub-sequences, let number them as 1st sub-sequence and 2nd sequence. So let first 0 is belong to 1st sub-sequence, then 2nd 0 should obviously belong to 2nd sub-sequence. Now, first 1 can either be belong to 1st sub-sequence or 2nd, and the last 1 would be in the alternate sub-sequence. So result can either be 1 2 2 1 or 1 2 1 2 or 2 1 1 2 or 2 1 2 1. I think you have got this now, right?
 » 4 years ago, # |   0 Regarding problem F, I can't figure out how does it mean. The first example↓, why the output is 3? In my understanding, it should be 1, because only (2,3) or (2,4) can be chosen. Otherwise, we will have the intersecting, ex: (2,4) & (2,3) have intersecting.  input: 4 1 5 2 4 2 3 3 4 output: 3  Is there anyone can help? THANKS.
•  » » 4 years ago, # ^ |   0 Your task is to choose the maximum by size (the number of segments) subset of the given set of segments such that each pair of segments in this subset either non-intersecting or one of them lies inside the other one.what does this statement mean? so confused. ><
 » 4 years ago, # |   0 This test in E1 problem 10 28 8 2 8 5 1 4 6 1 10 10 2 7 7 2 1 9 2 1 2 1 5 4 1 9 3 2 5 Why output is 7 ?
•  » » 4 years ago, # ^ |   0 10 28 8 2 8 5 1 4 6 1 10 10 2 7 7 2 1 9 2 1 2 1 5 4 1 9 3 2 5 Sorry i dont know how to end line :((
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 Can someone please explain how the answer for this test case is 7 instead of 8??The methodology I seem to get is :- edge 1-2 edge 6-1 edge 1-2 edge 1-4 edge 2-8 edge 2-10 edge 1-6 edge 2-3 Can someone explain as to how is it possible to do any better than this ? Thanks in advance!
 » 4 years ago, # | ← Rev. 2 →   +1 Can Anyone explain the proof of solution D? I didn't get the proof given in the editorial.
 » 4 years ago, # |   0 Can anyone tell will this work in E2?? In E2 can we maintain two multisets ,say one for overall costs and second only for edges with cost 1.Now we proceed similarly to E1 ,when we will encounter a edge with cost one we will simply include that in answer and continue. If we encounter a edge with cost two ,we will check for best two edges with cost 1 ,if we get better result than the cost 2 edge we will include thiese two otherwise cost 2 will only be taken.
•  » » 4 years ago, # ^ |   0 I think it will work. I am applying same thing but storing only edge with cost 1 in priority_queue and cost 2 in other priority_queue.
•  » » 4 years ago, # ^ |   0 Yes, It worked.
 » 4 years ago, # |   -7 RIP C :(
 » 4 years ago, # |   0 Can somebody tell me a test-case where my solution for D fails ? Link to solution : — https://mirror.codeforces.com/contest/1399/submission/89049595
•  » » 4 years ago, # ^ |   +1 Try this input: 1 12 110110001010 Correct output: 3 1 2 1 1 3 2 1 3 2 2 1 1 Your output: 4 1 2 1 1 4 1 2 3 1 1 1 1 
•  » » 4 years ago, # ^ |   0 I'm just trying to help. pks18 and Greatest789, in such cases, where you want to find a test case, you can use stress testing with an AC solution. Take a look at this blog
 » 4 years ago, # | ← Rev. 2 →   +1 Since yesterday I have been reading the question incorrectly. Until I saw the tutorial, I thought it was like that"Find minimum numbers of moves such that sum of weights of paths from root to every leaf is less then or equal S"Suppose root is a and leaves are x and y. Distance from a to x is p and a to y is q. p and q should be less then or equal to S but p+q can be greater then S. Example5 501 3 1001 5 102 3 1235 4 55ans is 7By the way, how to solve this?
 » 4 years ago, # |   0 Can someone explain what's wrong with my code? (Problem E1)https://mirror.codeforces.com/contest/1399/submission/89052647
 » 4 years ago, # | ← Rev. 2 →   0 I had used a max heap instead of ordered set but my approach was the same as that of the editorial. My solution gave TLE in Test 5.Is heap slower than set or did I do something wrong? Soln. Link: https://mirror.codeforces.com/contest/1399/submission/89023880
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 Heaps are actually faster then sets, your solution is giving TLE, as you are passing graph(vector g) by value, which increases the time complexity. Pass Graph(vector g) by reference or make it a global variable.
 » 4 years ago, # |   +3 can someone help me with 89027795 . I got TLE on test case 16 while in custom invocation its showing only 30 ms as well in my pc .
 » 4 years ago, # |   +3 I have a very different solution to E2 (no binary search, only linear DP). You can read about it here: SolutionI can try to explain and prove my solution ^_^Let's define impact of an edge as $(w-\left\lfloor\frac{w}{2}\right\rfloor)*l$, where $w$ is the current weight of that edge, and $l$ is the number of leaves in the subtree of that edge. So w occurs in sum $l$ times.Now let's find out, after performing, say, $i$ moves on edges with $c==1$ and $c==2$, the maximum sum of impact we can obtain. The resultant sum obviously decreases by sum of impact.Now, we do dp. $dp[i].first=max(dp[i−1].first+b1[dp[i−1].second.first+1],dp[i−2].first+b2[dp[i−2].second.second+1)$ where, $dp[i]=$ {maximum sum of impact for cost $i$, {number of moves with $c==1$ type edges, number of moves with $c==2$ type edges}}$b1[j]=$ maximum sum of impact with $j$ moves of $c==1$ type. Similar for $b2[j]$.Now an insight: The number of moves is bounded above by $nlog(nWmax)≤50n$ for the given constraints.So now we can just do linear search and output the answer just when sum$−dp[i].first$ becomes less than or equal to $S$.Even though DP relation is intuitive, here is the proof:Obviously, the Unable to parse markup [type=CF_MATHJAX] or $2$ type, so we max over those. Now, lets say for $1$ type, we do not take the $(dp[j−1].second.first+1)^{th}$ $1$-type move:UPD: A part of the previous proof was wrong. Here is the complete correct proof for the DP recurrence.Structure theorem: Let number of $1$ and $2$-type moves $(n1[i],n2[i])$. Then for adjacent elements$|n1[i]−n1[i−1]|≤1$$0<=n2[i]−n2[i−1]<=1$ (can be proved by easy induction)From here, we can easily compare the optimal solution and the solution of our DP+Greedy.
 » 4 years ago, # |   0 can someone tell me the testcase where my code is failing for E1well it is failing at 22nd case of 2nd pretest but it is not visible so please help me out
•  » » 4 years ago, # ^ |   0 Heyy, you need to use a custom comparator in your priority queue so that your top element is always the one that results in max decrease in the sum.class Compare { public: bool operator() (pair a, pair b) { int x = a.second/2; int y = b.second/2; return a.first * (a.second - x) < b.first * (b.second - y); } }; Here the 1st element of the pair is the no of leaves and the 2nd element is the weight.
 » 4 years ago, # |   0 hello everyone!Solutions of C given is in O(n^2). This can question can be solved in better complexity? If you have any idea suggest
•  » » 4 years ago, # ^ |   -8 No, I don't think so. We positively need to iterate over all possible values of sum, and for each sum, we need to find how many groups we can make. And after that, we'll find the maximum number of teams. So, the order has to be O(n^2).
 » 4 years ago, # |   0 how to install c++17 in local computer
•  » » 4 years ago, # ^ |   0 If you're in Linux and you have the appropriate GCC version (>5), you can just use the flag -stdc++=17 to use c++17.
•  » » » 4 years ago, # ^ |   0 no i'm a window user
•  » » » » 4 years ago, # ^ |   0 Install a higher version of Mingw on windows. The lower versions don't support C++17.
 » 4 years ago, # |   0 For Problem F, why the following greedy algorithm is not correct (WA on test 2): For every segment, calculate the number of trouble it makes to others (i.e. the number of segment that it intersects with) sort descendingly by trouble while !(everyone's trouble == 0) remove the segment that caused most trouble answer equal to the number of remaining segments
 » 4 years ago, # |   0 problem D why my submission giving sometimes TLE and sometimes WA. https://mirror.codeforces.com/contest/1399/submission/89094614
•  » » 4 years ago, # ^ |   0 in : 1 5 00100 out : 2 1 2 1 1 2 you can see this test case says it all. 2 set was starting from 0 and your code is giving the next element as 0 also in the second subsequence.
•  » » » 4 years ago, # ^ |   +1 yup I got it thanks
 » 4 years ago, # |   0 here is an alternative solution to problem C, simple brute force for all possible s, and then using 2sum to calculate. #include #include #include using namespace std; int main () { ios_base::sync_with_stdio(false); cin.tie(NULL); int tt; cin >> tt; while (tt--) { int N; cin >> N; vector A(N); for (int i = 0; i < N; i++) cin >> A[i]; int ans = 0; sort(A.begin(), A.end()); for (int k = 2; k <= N * 2; k++) { int i = 0, j = N - 1; int tans = 0; while (i < j) { if (A[i] + A[j] < k) i ++; else if (A[i] + A[j] > k) j --; else { tans ++; i ++; j--; } } ans = max (tans, ans); } cout << ans << '\n'; } return 0; } 
 » 4 years ago, # |   0 vovuh Can you look into my submission where it is giving WA or can you provide me a test case where it is failed?
 » 4 years ago, # |   -8 Easiest DIV3 Ever
•  » » 4 years ago, # ^ |   +2 Yes easiest when you cheat and your solution gets skipped ;)
•  » » » 4 years ago, # ^ |   0 Nice reply LoL.
 » 4 years ago, # | ← Rev. 2 →   0 For Problem E1 testcase 1 10 23 8 7 5 2 1 9 10 5 6 6 5 9 5 3 5 3 2 2 4 3 8 7 6 2 9 8 7 How does this can be done in just 12 steps, My solution divided following edges: total edge Value of particular edge(Column 1) no. of Leaves of going through that edge(COLUMN 2) Total Sum of paths(COLUMN 3) Number of steps(COLUMN 4) 80 27 3 65 -1 12 3 59 -2 10 2 53 -3 9 1 48 -4 8 1 44 -5 7 1 40 -6 6 3 37 -7 6 3 34 -8 6 1 31 -9- 5 1 28 -10 4 2 26 -11 4 1 24 -12 4 1 22 -13 Total steps= 13 given answer=12
•  » » 4 years ago, # ^ |   0 Maybe you are not taking the difference as the factor to decide which edge should be halved first.
 » 4 years ago, # |   0 Sorry if this is a stupid question, but can anyone tell me if my method of finding leaves is incorrect. I am just trying out all those elements whose adj[u].size()==1 Here is my function input, and check - vector adj(n+1); for(int i=0; i>v>>u>>w; adj[v].pb(u); adj[u].pb(v); } for(int i=1; i<=n; i++) { if(i!=1&&adj[i].size()==1) { leaves.pb(i); } }
•  » » 4 years ago, # ^ |   0 For n=2, it will consider both the vertices as leaves. In fact, for any tree where the root has only 1 degree will have it as a leaf. Ask yourself if that's what you really want or whether this makes a problem in your implementation (whatever you want to do with the leaves). You can start from 2 as you know that 1 is root. But yes, other than this corner case, your logic is ok.
 » 4 years ago, # | ← Rev. 2 →   0 Can anyone explain what is wrong in this prob-D I am getting wrong on (tc 70) of 2 test case:- https://ideone.com/yXMA6y
•  » » 4 years ago, # ^ |   0 Same problem. I am getting answer as 4. While the jury's answer is 3. https://mirror.codeforces.com/contest/1399/submission/89116568
 » 4 years ago, # |   0 If anyone need detail explanation for D Here
 » 4 years ago, # |   0 The time limit for D is 2s, so can O(N^2) solution work for D? My solution for D
•  » » 4 years ago, # ^ |   0 No, O(n^2) would work if n was around 10^3.
•  » » » 4 years ago, # ^ |   0 But the time limit for D is 2s, so shouldn't the cap be 10^10 approx?
•  » » » » 4 years ago, # ^ |   0 Not at all. Your logic is correct if 2*(10^5) = 10^10 . lol
 » 4 years ago, # |   0 Can someone explain the meaning of this line of code given in the solution code for problem D. What is auto [to,id]? Are we iterating over pairs corresponding to g[v]? for (auto [to, id] : g[v]) { if (id == p) continue; dfs(to, id); if (p != -1) cnt[p] += cnt[id]; }
•  » » 4 years ago, # ^ |   0 Yes, we are iterating over pairs corresponding to g[v].
»
4 years ago, # |
Rev. 3   -17

I couldn't solve C because I was searching for the value of S rest of the time but I wrote my algorithm well...I didn't read the limit of the problem constraints well and feeling bad about it when it was accepted after just adding an extra loop now.....

# include <bits/stdc++.h>

using namespace std; typedef long long ll; typedef vector vi; typedef deque di; typedef pair<int,int> pi;

# define FOR(i,a,b) for(ll i=a;i<b;i++)

int main() { //freopen("contest.txt","r",stdin); //freopen("contest.txt","w",stdout); IOS ll t,n; cin>>t; while(t--) { cin>>n; ll a[n]; FOR(i,0,n) { cin>>a[i]; } sort(a,a+n); ll sum=0; ll best=-1; for(int index=2;index<=2*n;index++){ ll l=0,r=n-1; while(r>l) { if((a[l]+a[r])>index) r--; else if((a[l]+a[r])<index) l++; else { sum+=1; l++;r--; }

}  best=max(best,sum);
sum=0;
}
cout<<best<<endl;
}

return 0;

}

Spoiler
Spoiler
•  » » 4 years ago, # ^ |   +1 In a fair world you would have got 100 extra penalty for bad formatting.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 Actually after pasting the code the fonts have changed like that along with the spaces and lines.Why is my format not good?
•  » » » » 4 years ago, # ^ |   0 There are several options. Above the editor box there are some symbols, click them. A useful one is called "Spoiler", another one "Block". This is a spoilerInitially hidden text. One can use formatting here, too. cool code ... blah 
 » 4 years ago, # |   +5 vovuh Can you please explain about the time complexity of your solution of F?
 » 4 years ago, # | ← Rev. 3 →   0 In problem 1399C - Boats Competition, the valid range for the smallest distinct weight $i$, where $i < s-i$, could have been written as $[\max(1,s-n),\lfloor\frac{s+1}{2}\rfloor-1]$. 89178688
 » 4 years ago, # |   0 In question E1: Why dividing the max w[i]*cnt[i] does not work?
•  » » 4 years ago, # ^ |   +1 hey firstly nice thinking, but it will give incorrect output as we are not just dividing weight but also finding its floor.here is an example.Lets w[1]=2, cnt[1]=11 , and w[2]=5 , cnt[2]=4;according to your approach: (2*11)-22/2=11, (5*4)-20/2=10 , as first one is greater therefore we will use it.But actually:w[i]*cnt[i]- floor(w[i]/2)*cnt[i] =>(2*11)-(1*11)=11, (5*4)-(2*4)=12here actually second one is greater. hopefully you understand, and got the point.
 » 4 years ago, # | ← Rev. 2 →   0 if my code was passed during the contest how can u give tle later? i knew a different method with sets which would have passed but u gave me AC which should have stayed as it is. giving WA after the hacking period is totally ok but TLE is not pls look into it
 » 4 years ago, # |   0 Initially I got TLE in E1, after long long(hehe) search it turned out that I was(by mistake) using pair instead of pair in my PQ and looks like it's responsible for a huge runtime increase(900ms vs TLE). Is just type conversion responsible for this?AC: 89159051TLE: 89159067only difference between these two is PQ data type
•  » » 4 years ago, # ^ |   0 Hint: 89162649 is your submission resubmited with an assert. Can you see what the problem is?
•  » » » 4 years ago, # ^ |   0 Thanks for response,I can now see that 0's coming out of PQ are resulting in TLE but I'm not sure whether it's a coincidence that they are 0's and not some random negative numbers as is usually the case with overflows.
•  » » » » 4 years ago, # ^ |   0 Hint: 89166892Try to guess why the 0's are coming. Experiment.
 » 4 years ago, # |   0 Can D be solved without any extra space pos0 and pos1?
 » 4 years ago, # | ← Rev. 2 →   0 I think the time complexity in E1 and E2 needs checking. The heap (or any other container) in E1 has only $n$ elements, corresponding to all the nodes of the tree, and the loop runs $O(nlogw)$ times. So the complexity should be $O(nlogw logn)$. In E2, though, the heaps DO have $O(nlogw)$ elemnts, and they run $O(nlogw)$ times as above. So the complexity is $O(nlogw log(nlogw))$.When I started writing this comment, the complexity in E1 was wrong, E2 was correct. Now I see that E1 has been fixed, AND copied on E2 as well which made E2 incorrect. Am I missing something?
•  » » 4 years ago, # ^ |   0 The current complexity for both problems is correct. You have a good explanation for E1, and the explanation of E2 is obvious too. We run the same simulation as in E1, which runs in $O(n \log{n} \log{w})$ (because the worst case is the same and it does not depend on edges costs) and then do two pointers on arrays, not heaps. These arrays really contain $O(n \log{n} \log{w})$ elements both but we process them in linear time of their length.
•  » » » 4 years ago, # ^ |   0 Oops. I forgot about the 2 pointer part. I was just mimicing the E1 solution. Thanks.
 » 4 years ago, # |   0 In solution for problem F Yet Another Segments Subset if you modify the lineif(l>r) return dp[l][r];to if(l>r) return 0;and make it the first line of cal(int l,int r)then we will not need those ternary operators later on!!
 » 4 years ago, # |   0 Can someone tell me why 1LL was multiplied in the problem E1?
 » 4 years ago, # |   0 Can someone please explain solution of question F? I am not able to understand it after so many tries.
 » 4 years ago, # |   0 Problem E2: In the 3rd test case of the input:2 1001 2 409 2Why is the answer "6"?For 1st coin, 409 -> 204 -> 102 ->51 . This takes 3 steps and similarly, 3 steps for the 2nd coin gives 51 and summing both give 102 which is greater than 100.Can anyone explain what am I missing?
 » 4 years ago, # |   0 Why is my solution using multisets giving tle(on test case 2) in problem E(easy version)?Using priority queues is NOT helpful either as it is giving tle on test case 5.
»
4 years ago, # |
0

D using Binary Search

Spoiler
 » 4 years ago, # |   0 What is the purpose of the line:cerr << cur << endl;In the solution of E1?
 » 4 years ago, # |   +1 I saw Geothermal got some memory issues because of using mapHe was not the only contestant!
 » 4 years ago, # |   0 Hi! For problem E2, my idea is: for edges cost 2 coins, it should be more expensive than the edge with similar weight to it, but cost only 1 coin. So I tried to maintain a set, which contains {g(i) , i} , where g(i) = (3 — c[i]) * (w[i] * cnt[i] — w[i] / 2 * cnt[i]) , where 1 <= c[i] <= 2 , w[i] being the weight of the i-th edge, and cnt[i] being the number of leaves under the subtree of the i-th edge. What my code does is, whenever the current sum is greater than S, we should take the largest pair in the set, sum -= (w[top] * cnt[top] — w[i] / 2 * cnt[top]) and erase {g(top) , top} update w[top]:= w[top]/2, and re-insert {g(top) , top} into the set. After implementing this greedy, I failed at test case 21. I wonder why this greedy fails, if anyone happens to be able to give a counter-example to my approach, please let me know! My implementation: https://mirror.codeforces.com/contest/1399/submission/90381838
•  » » 4 years ago, # ^ |   0 I think I figured out. if we can finish in reduction of 3, but 2 * 2 > 3 * 1 this is the counter example
 » 4 years ago, # |   0 Can someone help me to find what's wrong in my code(I also provided a little bit explanation in the code) : 90432879
 » 4 years ago, # |   0 https://mirror.codeforces.com/contest/1399/submission/93405575i got TLE on tc 6, but i thought it was O(N log N) , any one can tell me where's my fault
•  » » 4 years ago, # ^ |   0 I'm pretty sure it has got to do with the use of erase(). A comment above mentioned this: if you have an array of n length and you erase at index 0, you will have to move all n-1 elements to the left by one. This is perhaps why there's TLE.
 » 3 years ago, # | ← Rev. 2 →   0 I don't understand how the time complexity is $O(n^2)$ in F. Can someone please tell me how the total number of transitions is $O(n^2)$. Thanks in advance. vovuh
 » 3 years ago, # |   0 In problem E2 it is also possible to just fix parity of the answer and solve it separately for odd and even result, and in each turn adding a cost of 2 either by the best with the cost of 2 or by two bests of a cost of 1. My code is not the prettiest but just in case I will leave a link to the submission.
 » 14 months ago, # | ← Rev. 2 →   0 Way simpler approach for problem F:Do coordinate-compression on bounds of segments. We then iterate over segments in increasing order of size. For each segment, we will run O(n) computation which will find the answer for the problem if the required range was that of the segment.How to compute answer for each segment? Just sort segments by right border and store dp[r] = maximal number of segments we can get in [currentSegmentLeftBound, r]. Transitions are also very simple. Suppose we try to include segment s into selection, transition is dp[s.R] = max(dp[s.R, dp[s.L — 1] + score[s]).Implementation: link