### atcoder_official's blog

By atcoder_official, history, 4 weeks ago,

We will hold Toyota Programming Contest 2024#7（AtCoder Beginner Contest 362）.

We are looking forward to your participation!

• +38

 » 4 weeks ago, # |   -53 Can you give the topics for each question
•  » » 4 weeks ago, # ^ |   -51 A-Brute ForceB-G- Segment Tree and Binary search problems.
 » 3 weeks ago, # |   -37 Good luck, my friends!
•  » » 3 weeks ago, # ^ |   -37 you,too!
 » 3 weeks ago, # |   -18 GL&HF!
 » 3 weeks ago, # | ← Rev. 2 →   -10 ..
 » 3 weeks ago, # |   +22 Liked F but G ruined everything
•  » » 3 weeks ago, # ^ |   +5
•  » » » 3 weeks ago, # ^ |   0
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +3 Damn that's just sad now
•  » » » » 3 weeks ago, # ^ |   0 But it doesn't demand a certificate.
•  » » » » 3 weeks ago, # ^ |   0 This is very similar too.
•  » » » 3 weeks ago, # ^ |   -8
•  » » 3 weeks ago, # ^ |   +10 how did u do F was it Hungarian?
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   +22 My logic:Let the root of the tree be 1, you will notice that the most optinal way to the make the matchings is to make pairs between nodes such that they are in subtrees of different children of 1.How you make the matching doesn't matter since the total weight of the matching=sum of distances of all nodes from root i.e. 1.Now, if any children of the root of the tree i.e. 1 have subtree size > $n/2$ where n is number of vertices in tree. You can't make matchings for all the nodes in the optimal way mentioned above. To overcome this, reroot the tree such that the subtree size of all the children of the root is <= $n/2$. This root always exists and can be found using dfs.Now, To implement the construction of the matching there are many ways. What i did was make a set of the subtree sizes of children of the root and then take a node from both the subtree with smallest and largest size and pair them up, it doesn't matter what node you take as long as each node is taken once. Do this till all nodes have been matched.
•  » » » » 3 weeks ago, # ^ |   0 I just find that I have almost the same idea as yours (but failed finishing it during contest), thank you so much for sharing your solution! By the way, in the step of "reroot the tree ...", is the new root just the centroid of the tree? (I think it is but not quite sure)
•  » » » » » 3 weeks ago, # ^ |   0 Yes, it is the centroid
 » 3 weeks ago, # |   0 E took my energy
 » 3 weeks ago, # |   +43 Are you kidding? Why is task G the template of the Aho-Corasick Automaton?Here is the template task in Luogu. They are the same.
•  » » 3 weeks ago, # ^ |   0 and like this: https://vjudge.net/problem/LightOJ-1427
 » 3 weeks ago, # |   +6 Is g some standard problem?
•  » » 3 weeks ago, # ^ |   +6 Yes, very standard. Either Aho or Suffix Array will do it.
•  » » 3 weeks ago, # ^ |   +5 yep its a cses problem
•  » » 3 weeks ago, # ^ |   0
•  » » 3 weeks ago, # ^ |   +3 How could you not know that with that username
•  » » » 3 weeks ago, # ^ |   0 I know trie but not aho corasik algo
 » 3 weeks ago, # |   +40 Please stop giving problems that can be solved by just copying the template like G.
 » 3 weeks ago, # |   0 how to solve E
•  » » 3 weeks ago, # ^ |   0 very ez that dp
•  » » 3 weeks ago, # ^ |   0 I did it using 3d dp.This is my submission :https://atcoder.jp/contests/abc362/submissions/55562909 (was my first contest at atcoder :))I set dp[i][length][diff] = number of such sequences such that the ending position of subsequence is i , the length of subsequence is length and the difference of the AP is diff , i kept diff in a dictionary(map) as the value of diff could reach upto 10^9.After that just used dp[i][length+1][diff] += dp[j][length][diff]Since if there is subsequence of length k with difference diff and the current element minus j th element is equal to diff , then we could make a new subsequence of length legth+1Then for the final answer , just printed the summation of length for all i and diff.
 » 3 weeks ago, # |   0 How do you solve C? I managed to solve D and E but couldn't solve C
•  » » 3 weeks ago, # ^ |   0 The minimun sum u can construct is sum of all l[i] and maximun is sum of all r[i]
•  » » 3 weeks ago, # ^ |   0 lmfao I wish I skipped C then coz by the time I finished it I had no time for either D or E
•  » » 3 weeks ago, # ^ |   0 Yes,me too.D is a standard problem and E is a DP problem.But C is harder than D.
•  » » 3 weeks ago, # ^ |   0 To construct the Xs you can initialize them to the maximum sum they can give, then decrease them until the sum is 0. You can also probably go in the reverse direction as well, start with minimum sum and increase
 » 3 weeks ago, # |   0 Can't believe so many people would SAM.
•  » » 3 weeks ago, # ^ |   0 In fact,ACAM is enough.
•  » » » 3 weeks ago, # ^ |   0 Perhaps a more appropriate statement would be "KMP on Trie".
•  » » » » 3 weeks ago, # ^ |   0 Also can't believe so many people would ACAM. ==
•  » » 3 weeks ago, # ^ |   0 You can also use ACAM to solve G.
•  » » 3 weeks ago, # ^ |   +7 Most of them just copy the others' code. There is a similar problem on a Chinese OJ and there are published solutions with code.
 » 3 weeks ago, # | ← Rev. 3 →   +100 Atcoder, if you aren't able to set Gs then just remove them.
 » 3 weeks ago, # |   0 Solved F a couple of minutes after the contest has ended, eternal pain and sufferingShould have taken a hint from the problem name. Btw what's the actual way to do a matching in a case like this, without bruteforcing and abusing move-semantics?
 » 3 weeks ago, # | ← Rev. 2 →   +29 Problems such as problem G shouldn't appeared in the contest. It is too original, and you just need to copy and paste without any thinking. Even worse to make G 575 but F 550, it should be reversed. It's super unfair for candidates who can solve F but doesn't know anything about AC automation.
•  » » 3 weeks ago, # ^ |   -10 Don't know ACAM means too weak. Just practise more.
 » 3 weeks ago, # | ← Rev. 2 →   0 how to do F ?I could get somewhat idea to use Hungarian but that O(n^3) then whats the way to solve F or am i overthinking and using something which is useless like Hungarian in this case
•  » » 3 weeks ago, # ^ |   +3 I kind of guessed F, if you find a centroid and start matching nodes from different subtrees of centroid then the actual pairs do not matter since they won't affect the answer, ie sum of all distances should stay the same (at least it feels like it). Then you have reduced the problem to matching integer values from different sets.
 » 3 weeks ago, # |   +8 I don't like this round. It has a template problem like G, it makes this round boring and unfair.(just my opinion)
 » 3 weeks ago, # |   0 Solved E after 3 mins the contest ended... Fortune enough to notice G is suffix array to cover the loss in E.
•  » » 3 weeks ago, # ^ |   0 Can you tell me why this code fails I have a counter case I am not able to figure out where I am wrong 5 1 1 1 1 1(Code)
 » 3 weeks ago, # |   +3 E is similar to this problem.
•  » » 3 weeks ago, # ^ |   0 But their ideas are different.
 » 3 weeks ago, # |   +1 The G problem is not a original problem.
 » 3 weeks ago, # |   +9 Delete G.
 » 3 weeks ago, # | ← Rev. 2 →   0 Can anyone help I am getting tle on last 2 test cases of problem d int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); lli n,m; cin>>n>>m; vectorarr; arr.push_back(0); for(int i=0;i>x; arr.push_back(x); } vector>mp[n+1];for(int i=0;i>x>>y>>z; mp[x].push_back({y,z}); mp[y].push_back({x,z}); } vectordis(n+1,1e18); dis[1]=0;priority_queue, vector>, greater<>> pq; pq.push({arr[1],1}); while(pq.size()>0){ lli dd=pq.top().first; lli vv=pq.top().second; pq.pop(); for(auto it:mp[vv]){ if(dis[it.first]>dd+it.second+arr[it.first]){ dis[it.first]=dd+it.second+arr[it.first]; pq.push({dis[it.first],it.first}); } }} dis[1]=0; for(int i=2;i<=n;i++){ cout<
•  » » 3 weeks ago, # ^ |   0 bhai, you are pushing same node multiple times in the PQ. Lets say, graph, lets say, the input is 4 5 10 10 10 10 1 2 20 2 3 1 1 3 1 1 4 1 4 2 1 Please check, how many times you are pushing node value '2' in the queue.
•  » » » 3 weeks ago, # ^ |   0 Yeah I got my mistake,thanks a lot
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 You should create an array called bool vis[] which stores whether each node has already been processed. Without this array, sometimes a node can be processed multiple times. Although this doesn't result in an infinite loop, this still significantly adds to the execution time. (Processed multiple times means the neighbors of the node is relaxed multiple times, which isn't necessary.)FYI the solution to this problem is that for every edge u->v with weight w, we simply add arr[v] to w. Then, we run a standard dijkstra and add arr[1] to our answer. Proof of correctness is fairly straightforward.PS: It is recommended to send submission links rather than the whole code. This is not only more aesthetically pleasing but also gives us more information on the submission, whether it TLEs or WAs, etc.
•  » » » 3 weeks ago, # ^ |   0 Thanks I got my mistake, yes I'll paste the link next time onwards
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 can you share the submission link using visited array with this code ?
•  » » » » » 3 weeks ago, # ^ |   0
•  » » » » » » 3 weeks ago, # ^ |   0 yeah, it works same as using the visited array before checking for it's neighbours, if the node is visited then continue
 » 3 weeks ago, # |   +11 I copied pasted my solution to G from CSES and changed 0 lines. I don't think problems like this should appear in a rated round. If problems like this are to appear, it should be in an unrated educational round (similar to the Atcoder DP contest a long time ago).
 » 3 weeks ago, # | ← Rev. 3 →   -10 [DELETED]
 » 3 weeks ago, # |   +6 This round was particularly frustrating, because I haven't seen G before and therefore spent a substantial amount of time on G. Hence, I finished the contest with ABCDE_G. However, I realized the greedy solution for F merely 5 minutes after the contest. If I had seen G before, I could focus on F and would have definitely full-solved. Not only did I not full-solve, my rating is probably going to be affected too, due to the fact that so many people have seen G before. I think many people are having the same thoughts, so atcoder_official, if you can't set G's, then don't. Having a contest with 6 problems is better than having a template question as the highest-point problem.
•  » » 3 weeks ago, # ^ |   0 Can you please check this submission ? why is this getting WA, just can't figure out. Can't figure out the error :| . https://atcoder.jp/contests/abc362/submissions/55580896jatrophalouvre
•  » » » 3 weeks ago, # ^ |   0 Breakcase: 10 13 1660 7612 4840 3444 6628 7079 9647 9324 803 9135 8 2 6893 4 4 8584 10 5 4386 2 5 9772 10 8 3026 10 9 2303 9 5 3584 7 8 654 9 10 208 6 3 1571 1 2 6613 1 3 2485 1 4 9067 Your code gives: 15885 8985 14171 32285 17635 42403 32102 36672 46015 Correct answer: 15885 8985 14171 32285 17635 42403 32102 36672 44263 
•  » » » » 3 weeks ago, # ^ |   0 thanks a lot.
 » 3 weeks ago, # |   +55 In fact, problem F is so similar to 1387B2 - Village (Maximum).
 » 3 weeks ago, # |   0 can someone please help me figure out the mistake in this ? https://atcoder.jp/contests/abc362/submissions/55580896
 » 3 weeks ago, # |   +8 Does Dijkstra with Min Heap priority queue not work for anyone for D? It gives TLE for me. So is the solution using Fib Heap? Or is it some optimization issue?
•  » » 3 weeks ago, # ^ |   +8 It worked for me with just some slight modification - At line 52, 53 and 69, 71 --> rest the whole algorithm will be normal djikstraYou can check — My submission
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   +8 Ohh it looks very similar to my approach too!Can you check this submission out: https://atcoder.jp/contests/abc362/submissions/55586366I'm not sure of the TLE in this. The approach is more or less the same. Also in my previous version of reply I had a silly mistake of not including V,U edge which caused WA
•  » » » » 3 weeks ago, # ^ |   +8 Inside the solve function --> you are creating a self loop => which may cause both TLE and Wrong answer graph[u].push_back(Edge(v, b)); graph[u].push_back(Edge(u, b)); --> v must be thereI think it should get accepted, feeling bad for the typo..
•  » » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +8 Lol yea saw that. That was silly haha. Still TLE tho.Edit: Turns out my visited array I was using for not doing repetitive computation was not doing the trick. I should've used the dist array itself. Got it now. Thanks for helping me out bro!
 » 3 weeks ago, # | ← Rev. 2 →   0 Problem G is a template of Aho–Corasick algorithm.
•  » » 3 weeks ago, # ^ |   +3 By the way, it's not Aho–Corasick algorithm but Aho–Corasick Automaton.
•  » » » 3 weeks ago, # ^ |   0 But it's such a name in Wikipedia.
•  » » » » 3 weeks ago, # ^ |   0 Well, you see,Aho-Corasick is not a Algorithm but a data structure based on trie, so I think there might be an error in Wikipedia.
•  » » » » » 3 weeks ago, # ^ |   0 Yes you are right, its name is indeed this.
 » 3 weeks ago, # |   0 Hello there!. Please help me with this submission why TLE!!https://atcoder.jp/contests/abc362/submissions/55595549
•  » » 3 weeks ago, # ^ |   0 Hi!You should add if (vis[u]) continue; to 145th line.
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Yeah I got it. ThanksBut why exactly this works. You can see that I am checking for vis while traversing for all the neighbors.
•  » » » » 3 weeks ago, # ^ |   0 Possible reasons for this are that a point's neighbor is placed on the back end of the priority queue, causing the point to not be marked in time, and it can also cause a node to queue many times and time out.
•  » » » » » 3 weeks ago, # ^ |   0 Ok I see. Traversing the neighbors of the same node more then once is a problem. If the node enters in the pq then it will get it neighbors checked this may happen multiple times. Now I get it.Thanks mate!!. (Mentioning this to not follow the same in future).
 » 3 weeks ago, # | ← Rev. 3 →   0 Can anyone please spot the mistake in my code for C question I am getting 2 Wrong answers rest all accepted.My submission: https://atcoder.jp/contests/abc362/submissions/55602146
•  » » 3 weeks ago, # ^ |   0 Try this input: Spoiler2 1 5 -10 -5 
•  » » 3 weeks ago, # ^ |   0 Fixed in this submission
•  » » » 3 weeks ago, # ^ |   0 thank you loll forgot to print the answer itself
 » 3 weeks ago, # |   0 t o y o t a
 » 3 weeks ago, # | ← Rev. 2 →   0 Edit — solved
 » 3 weeks ago, # |   0 how come my O(n^5) code is running for problem e https://atcoder.jp/contests/abc362/submissions/55674291 I think it shouldn't... someone please correct me if i am wrongly calculating complexity or the constraints are just too loose
 » 3 weeks ago, # | ← Rev. 2 →   0 Can anyone explain me the approach Jinagly used for problem G. https://atcoder.jp/contests/abc362/submissions/55526066 What does the array fail indicate?
•  » » 2 weeks ago, # ^ |   0 it is fail pointer in Aho-Corasick algorithm