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By awoo, history, 5 years ago, translation,

Tutorial

1197B - Pillars

Idea: BledDest

Tutorial
Solution (BledDest)

1197C - Array Splitting

Idea: Roms

Tutorial
Solution (Roms)

1197D - Yet Another Subarray Problem

Idea: BledDest

Tutorial
Solution (Roms)

1197E - Culture Code

Tutorial

1197F - Coloring Game

Idea: BledDest

Tutorial
Solution (BledDest)
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 » 5 years ago, # | ← Rev. 2 →   +40 I challenge you to write D in $O(n)$ SolutionFirst of all if $m \geq n$, let's make $m = n$.Let's us look at some segment $[l, r]$. The actual cost of it is $\sum_{l}^{r} a_i - \left \lfloor{\frac{r - l + 1}{m}}\right \rfloor \cdot k - (k$ or $0)$ depending on the segment. Let's subtract $k$ from each $a_{m \cdot i}$. Now we can be sure that the cost of $[l, r]$ is $\sum_{l}^{r} a_i - (k$ or $0)$ depending on the segment. How to differ those cases? Let's look at a half-interval $(l - 1, r]$. $\sum_{l}^{r} a_i$ is equal to $prefsum_r - prefsum_{l - 1}$. How do we define whether we should subtract $k$ or not? Let's look at $r \pmod m$ and $l - 1 \pmod m$. If $l - 1 \pmod m \geq r \pmod m$ than we alredy subtracted the right amount of $k$'s. Otherwise we should subtract another $k$.Let's write it in $O(n \cdot m)$. Let's maintain $m$ values $mn_i =$ minimal prefsum of some index $j$ such that $i \equiv j \pmod m$. Now let's run from $0$ to $n - 1$ with index $i$. On each iteration let's consider all segments such that right bound is $i$. Now run through $mn$ and and update our answer if we could. After that let's make $mn_{i \pmod{m}} = min(mn_{i \pmod{m}}, prefsum_i)$.Now how to make it $O(n)$. let's notice that answer for $i$'th index is $prefsum_i$ — $min(min(mn[0..i$ $\%$ $m - 1]) + k,$ $min(mn[i$ $\%$ $m..m - 1]))$. So basically we need suffix minimum, prefix minimum and update one value (the first value in current suffix). Let's process each block of length m separately. For each block let's update suffix minimum in $m - 1$ updates. Let's define prefix minimum as $p = \infty$. Now while running through $i$ to $i + m$ let's update $mn_{i \pmod{m}}$ with $prefsum_i$ and after updating answer with $prefsum_i - min(p + k, suffmn_{i \% m})$ update $p$ as $min(p, mn_{i \% m})$. On each block it takes $O(m)$ iterations, and there are at most $\left \lceil{\frac{n}{m}}\right \rceil$ blocks. So it's $\left \lceil{\frac{n}{m}}\right \rceil \cdot O(m) = O(n)$, as $m \leq n$.You can also notice that this solution doesn't need to store all the array, so the only memory it takes is $m$ values of $mn$ and other $O(1)$ values. So it's $O(n)$ in time and $O(min(n, m))$ in memory. This solution is the fastest of all and also uses 0KB of memory.UPD: you can also optimize the solution I've written below using queue minimum. But it's not as nice and causes bigger constant.
 » 5 years ago, # |   +3 What's the intuition behind D?
•  » » 5 years ago, # ^ |   +15 I can explain a different (easier) solution. Let's define $dp_i$ as answer, such that $i$ is the last used element. Now $dp_i = max(0, a_i - k, max(dp[j] - k + \sum_{j + 1}^{i} a_t)$, where $i - j \leq m$. The intuition is as follows: we need to divide all the segment, into subsegments, each length $\leq m$.
•  » » » 5 years ago, # ^ |   0 I used the technique but the it fails on test case 19 here is the link to it.
•  » » » » 5 years ago, # ^ |   +3 It's a little hard to understand your code. This is the implementation of what I wrote.
•  » » » » » 5 years ago, # ^ | ← Rev. 2 →   0 Hm that's kinda weird, cause when I tried to implement it here (https://mirror.codeforces.com/contest/1197/submission/57633664) I ran into problems with it applying the "k" penalty more times than it shouldIn the end I implemented a solution that uses a second "tolerance" value ( = (-len) mod m ) to indicate how many more values the subarray could take before the k penalty should be applied again: https://mirror.codeforces.com/contest/1197/submission/57643293 but now I'm puzzled as to why your code works without it! :PEDIT: Oh, I just realized that the first code sample had an issue where I wasn't getting the correct results for subarrays [0..i], where 1 < i < m . This bit me later in the butt in another test. The code now works without needing a tolerance term: https://mirror.codeforces.com/contest/1197/submission/57646083
•  » » » 5 years ago, # ^ |   0 Can you explain a bit more the meaning of each term in the recursive formula in terms of the decision it represents?
•  » » » » 5 years ago, # ^ |   +3 Basically we want to find the best segment, such that after splitting it into $cnt$ subsegments each length $\leq m$, $\sum_{l}^{r} a_i - cnt \cdot k$ is maximum. So when we want to count $dp_i$, we take all lengths $1$ to $m$ and try to make $[i - len + 1, i]$ the last of subsegments. That's how we update dp. This solution relies on the fact that $k \geq 0$.
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 length ≤m, ∑rlai−cnt⋅k is maximum. Can you plz explain thisEdit-ok i got that and you are very smart:)
•  » » » » » » 4 years ago, # ^ |   +3 I'm glad you did it by yourself =)
•  » » » » » » » 4 years ago, # ^ |   0 ༼ つ ಥ_ಥ ༽つ thanx
•  » » » » » » 4 years ago, # ^ |   0 Can you explain $best_i = \max\limits_{0 \le len, i - len \cdot m \ge 0} (sum(i-len \cdot m + 1, i) - len * k)$What is this the "best" of? How comes this formular?
•  » » » » » » » 4 years ago, # ^ | ← Rev. 2 →   +5 i will explain my solution which is similar to that solution (my sol https://mirror.codeforces.com/contest/1197/submission/80940806) 1.dp[i] will represent the subarray with max cost ending at i 2.now we will iterate from the last index 'i' up to when the length becomes m because the greatest integer of length less than m will be 1 so after that we don't have to because it will be same as dp[i-j] where j is the starting index3.now we will just maintain 'sum' of value let say we are at jth index from the last index i so sum =a[j]+a[j+1]...a[i] now we dp[i]=max(dp[i],sum-k) 'sum-k' because the length i-j+1 is less than or equal m also we have to do dp[i]=max(dp[i],dp[j-1]+sum-k)
•  » » » » » » » » 4 years ago, # ^ |   0 What is the key idea, or key observation to come up with that solution?
•  » » » 5 years ago, # ^ |   0 so for this solution, how does it take into account segments with length > m?
•  » » » » 5 years ago, # ^ |   +4 Every segment consists of subsegments of length $\leq m$. For each small subsegment you have to subtract $k$. So each segment is just continuous merge of small subsegments. That's why we can apply DP. Each segment is either a small segment or a merge of segment and a small segment.
•  » » » » » 9 months ago, # ^ |   0 why always subtract? let us say i=7, j=5,m=4 then ceil(i/m)=ceil(j/m).isn't it?
•  » » » » » » 9 months ago, # ^ |   0 Because the task is equivalent to all segments, which have the length of at most m. Our DP does just that, when we try to update it with last $m$ segments.
•  » » » 5 years ago, # ^ | ← Rev. 2 →   0 This easier solution is O(m*n) right?
•  » » » » 5 years ago, # ^ |   0 Yes
•  » » » 5 years ago, # ^ |   0 You can check my solution, Only $j = i - m$ needs to be considered, Otherwise just take max subarray ending at $i$ with length $\leq m$.
•  » » » » 5 years ago, # ^ |   +1 Yes, your solution is actually the same, I simply don't differ those cases. It still works in $O(n\cdot m)$.
•  » » » » » 5 years ago, # ^ | ← Rev. 2 →   +8 It can actually be improved to $O(n)$ easily by also storing the 2 best subarrays of length $\leq m$ for $i - 1$ so you don't have to do a for loop.
•  » » » 5 years ago, # ^ |   0 consider this case: say $j = i - 1, dp_j = a_j - k$, when you try to update the value for $dp_i$ using $dp_j - k + a_i$, $k$ will be deducted twice when you calculate $dp_i$. Is that the case or am I misunderstanding something?
•  » » » » 5 years ago, # ^ |   +3 Yes, in this case it subtracts twice, but we will find the best answer anyway.
•  » » » » » 5 years ago, # ^ |   0 Is it ok for you to explain why this is the case?
•  » » » » » » 5 years ago, # ^ |   0 I assume that [i, i] is the last small segment. It means that before i there were small segments only of length m. This leads to another subtraction as [i, i] is a different segment.
•  » » » » » » » 5 years ago, # ^ |   0 I mean why we can still find the optimal answer despite that k is subtracted twice
•  » » » » » » » » 5 years ago, # ^ |   0 This is the situation, where length is x * m + 1, we have to subtract k x + 1 times.
•  » » » » » » » » » 5 years ago, # ^ | ← Rev. 2 →   0 I kindda get your idea after re-reading your comments. In your method, although $dp[i]$ is miscalculated when we try to update it using $dp[i-1]$, the optimal answer will surface out when we try to update it using $dp[i-2]$, is that the case?
•  » » » » » » » » » 5 years ago, # ^ |   0 Yes, we choose the maximal value of all, so we will get the best answer.
•  » » » » » » » » » 5 years ago, # ^ |   0 I think I finally get it. Thanks for answering my questions patiently!
•  » » » 4 years ago, # ^ |   0 Thanks for clear explanation!
•  » » » 4 years ago, # ^ |   0 When you use the transition $dp[j] - k + \sum_{j+1}^{i} a_t$ , then you extend the optimal segment ending at $j$. How do we decide the $-k$ part here?
•  » » » » 4 years ago, # ^ |   0 Consider it this way: adding a segment of length $\leq m$ costs you $k$ money.
•  » » » » » 4 years ago, # ^ |   0 But may be when we add the segment, then the number of segments doesn't increase....Suppose $m = 5$ and $dp[j]$ comprises of segments of size $3,5,5,5$. If we use one more element to update $dp[i]$ then the segments will be $4,5,5,5$ . Here $-k$ won't be there.
•  » » » » » » 4 years ago, # ^ |   0 We will overlook both $4, 5, 5, 5$ and $3, 5, 5, 5$. I do not understand your usage of word "segment". We'll just try and add every last possible segment $[j+1, i]$ and update $dp[i]$, which is the best answer for all segments ending at position $i$.
•  » » » » » » » 4 years ago, # ^ |   0 In the above example with $m = 5$, I meant that since $dp[j]$ means the answer for position $j$ ,let's say optimal value is for the interval $[j - 17, j]$ which means value of $dp[j]$ is $\sum_{j-17}^{j} a_t - 4k$ . When we update something like $dp[j+1]$ using $dp[j]$ then we will take $dp[j] - k + a[j+1]$ , but it should only be $dp[j] - a[j+1]$ because now the interval becomes of size 19 and we have already paid $4k$ in $dp[j]$...
•  » » » » » » » » 4 years ago, # ^ |   +3 I think you just don't get the dp approach. Yes, we will overlook ALL possible decompositions and choose the best one. Please look at this thread: https://mirror.codeforces.com/blog/entry/68615?#comment-530099
 » 5 years ago, # |   0 I don't understand the solution for C, I guess this solution works because the initial array is sorted? Is there known algorithm that leads to this solution using differences between adjacent elements?
•  » » 5 years ago, # ^ |   0 https://mirror.codeforces.com/blog/entry/68553?#comment-529278 may be this comment will help u !!
•  » » 5 years ago, # ^ |   0 Let a[1], a[2].. a[n] be the elements in the array. Let 1 <= i < j < k < l....< n be the k-1 indices we choose for dividing the array into k segments. So, cost of division = (a[i] — a[1]) + (a[j] — a[i+1]) + .... + (a[k] — a[j+1]) + (a[n] — a[k+1]), on rearranging : (a[n] — a[1]) — ((a[i+1] — a[i]) + (a[j+1] — a[j]) +.....+ (a[k+1] — a[k])). So, we will store the difference of a[i+1]-a[i] for each 1<=i
•  » » 5 years ago, # ^ | ← Rev. 2 →   0 lol
 » 5 years ago, # |   0 Can someone explain how to solve E problem. Why do we need a segment tree. And how do we find the value of the minimum for the whole set of the dolls?
•  » » 5 years ago, # ^ | ← Rev. 3 →   +8 If there is no $j$, that $matryoshka_j.in \ge matryoshka_i.out$, then $d_i = (matryoshka_i.in, 1)$else$d_i.first = min[pos,n] - (matryoshka_i.out - matryoshka_i.in)$, where $pos$ is the first $j$, that $matryoshka_j.in \ge matryoshka_i.out$ (It means that we can put matryoshka $i$ inside matryoshka $j$)$d_i.second = numberOfMinimums[pos,n]$And $d$ — is our segment tree, so we need it to calculate minimum and number of minimums on suffix!You can check my solution, I think it's clear enought.
•  » » » 5 years ago, # ^ |   0 yep , your solution is self descriptive , but i have seen this https://mirror.codeforces.com/contest/1197/submission/57527419 which doesn't even use segtree , like no heavy DS stuff .
•  » » 5 years ago, # ^ |   0 I haven't mastered segment trees yet (did an object-pointer based one for practice but probably need to learn how to do it with bitwise-math and array indices for better efficiency) but I managed to solve this one using only a priority queue; code here if interested:https://mirror.codeforces.com/contest/1197/submission/57589994
 » 5 years ago, # |   +8 What does len mean in the editorial of D? And how does that expression len * m come?
•  » » 5 years ago, # ^ |   0 I suppose len is the number of segments that are fully covered. So $best_i$ is the answer if only segments of length divisible by m are considered.
 » 5 years ago, # |   0 I come with something for problem D but I can't develop it to a solution if anyone solves it by the help of the following please tell me:we deal with the array as the blocks, we try each possible one let's call it j from 1 to m, and for every j we try to start from 0 to j — 1 and move by j step to partition the array.it does nothing but I think it could be developed in some way.
 » 5 years ago, # |   0 Very nice explanation of problem E. Thank you so much)
 » 5 years ago, # | ← Rev. 2 →   +6 I was discussing problem E with my friend idgaf and after he solved it we read the editorial. And we noticed this part ( It's because not big enough subsets are not optimal in terms of minimality of extra space.) However, we are not sure if the part that adedalic says is true, please correct us if we're wrong. Take this example with 2 dolls {4, 1} {6, 5}The minimum extra space of big enough is 2, but if you take the first doll only, the extra space is 1
•  » » 5 years ago, # ^ |   +8 Let's extend your example as {2,1} {4,2} {6,5} $d[3] = (5, 1)$ — it should be obvious. When we try to calculate $d[2]$ we can see, that the second matryoshka can be nested inside the third one — so "we must put it inside". Then $d[2] = (5-2=3, 1)$.The first doll can be nested both in the second and third dolls, but putting it inside the third doll will lead us to not big enough subset. But! it also makes the extra space not optimal since $d[2].first < d[3].first$. Then $d[1] = (3-1=2, 1)$ and it's correct.
•  » » » 5 years ago, # ^ |   0 Ohh, I understand, I was a little confused, but now is totally clear. Thanks.
 » 5 years ago, # |   0 Can someone explain the intution behind D and as to how are we connecting the maximum sum subarray problem to this problem in greater depth. Also I'm not able to understand the intuition behind the idea of introducing len (as done in the editorial) Please help.
 » 5 years ago, # |   0 This is an alternate Solution 57537548 of Problem 1197C - Разделение массива
 » 5 years ago, # |   0 In problem E, Why cann't we add subset {4,7} as one of good subsets in Test Case 1.
•  » » 5 years ago, # ^ |   0 Because it is not big enough. We can put $7$ inside other doll
•  » » » 5 years ago, # ^ |   0 ok,got it.
 » 5 years ago, # |   0 can someone give me some further explanations about problem c? more specific: why should we "add up the k−1 minimal ones to the answer"?
 » 5 years ago, # |   0 In the solution given for D why do we fix some elements of best_i with best_i + sum(i-len,i-1) — k ?
•  » » 5 years ago, # ^ |   0 I think this is a typo: https://mirror.codeforces.com/blog/entry/68615?#comment-530337
 » 5 years ago, # | ← Rev. 2 →   0 Can someone explain in more details solution of C, with proofs? Thanks
 » 5 years ago, # | ← Rev. 2 →   +27 I don't see why we need a segment tree for E since all queries are suffix queries and all updates are at the position we are currently at. We can just maintain a suffix minimum as we go along.Solution with editorial's approach minus segment tree: 57644319
 » 5 years ago, # |   +3 So, I actually got hacked on problem A because of exceeded time limit. How do you know if your solution needs to be O(n) or O(nlogn) given the time constraint? Would you just assume 2 seconds means my solution needs to be O(n)? Sorry for too many questions, I'm just new to CP.
•  » » 5 years ago, # ^ |   +3 The rule of thumb I use is to plug the maximum input size into the big O, and look for something $\le 10^8$.So for $n=10^3$ for example, $O(n^2)$ is fine ($10^6 < 10^8$), but for $n=10^5$, it'd be too slow ($10^{10}>10^8$).In your solution you used quicksort, which is usually $O(n \log n)$, fast enough, but you were hacked by an input that causes it to become $O(n^2)$, which is too slow.
 » 5 years ago, # |   0 Could somebody explain me, if we want to calc T(i, j), we must know colors of r1, r2, r3. But I can't find any mention of it in the tutorial.
 » 5 years ago, # |   0 Logic behind soulution of C?
 » 5 years ago, # |   0 I think problem F can be solved simply by reducing the number of states. The number of states that are possible to reach is actually way lower than 64, it's 25. Then I don't think there's any need for optimization, just brute 25^3*1000*log2(10^9) will work. (Correct me if I'm wrong).
 » 5 years ago, # |   +3 In the tutorial for D, last line, you wrote: $best_i + sum(i - len, i - 1) - k$. But in the code posted below it seems that you are calculating $best_i + sum(i + 1, i + len)$. I think that is a typo in your tutorial.
 » 5 years ago, # |   +3 Can E be solved by Graph Theory? Just saw the "Shortest Path" tag.
•  » » 5 years ago, # ^ | ← Rev. 2 →   +3 There's an edge from i to j with length $in_i-out_j$ ,if and only if $in_i\ge out_j$ holds. Sort the matryoshkas by $out_i$ ,then we can use segment tree to build the graph. To solve the problem more easily,we use two more nodes,S and T. For all node i with indeg=0,add an edge from S to i with length 0. For all node i with outdeg=0,add an edge from i to T with length $in_i$ . The answer is the number of shortest paths from S to T. Since the graph is a DAG,the solution above works in $O(n\log n)$ time.
•  » » » 5 years ago, # ^ | ← Rev. 2 →   0 I got an WA on test 8 two days ago.My approach is similar to yours. 57926487
•  » » » 5 years ago, # ^ | ← Rev. 2 →   0 just found my mistake. Thanks anyway.
•  » » » » 5 years ago, # ^ |   0 AC code: 58050060
 » 5 years ago, # | ← Rev. 2 →   +10 Here is an easier $O(n \log m)$ solution for D: Let $p$ be the prefix sum of $a$. Then maximum cost of subarray ending at $i$ is $\max_{j < i} \left\lbrace p_i - p_j - k\left \lceil \frac{i - j}{m} \right\rceil\right\rbrace$Main observation is -- $\left\lceil \frac{i - j}{m} \right\rceil = \begin{cases}\left\lfloor \frac{i}{m} \right\rfloor - \left\lfloor \frac{j}{m} \right\rfloor + 1, & \text{if }(j \text{ mod } m) < (i \text{ mod } m) \\ \left\lfloor \frac{i}{m} \right\rfloor - \left\lfloor \frac{j}{m} \right\rfloor, & \text{if } (j \text{ mod } m) \geq (i \text{ mod } m) \end{cases}$So, our formula becomes -- $\begin{cases}p_i - k\left\lfloor \frac{i}{m} \right\rfloor + \left(k\left\lfloor \frac{j}{m} \right\rfloor - p_j\right) - k, & \text{if }(j \text{ mod } m) < (i \text{ mod } m) \\ p_i - k\left\lfloor \frac{i}{m} \right\rfloor + \left(k\left\lfloor \frac{j}{m} \right\rfloor - p_j\right), & \text{if }(j \text{ mod } m) \geq (i \text{ mod } m) \end{cases}$To evaluate the formula quickly, we can keep the maximum of $\left\lbrace k\left\lfloor \frac{i}{m} \right\rfloor - p_i\right\rbrace$ at index $(i\text{ mod } m)$ of a segment tree. Which enables us to get the prefix/suffix maximum in $O(\log m)$ resulting in total complexity $O(n \log m)$.
 » 5 years ago, # | ← Rev. 2 →   0 the second example of B: 3 1 2. why it is NO？we can move it like this.-------------------------1------------------1---------------1-------------------------------23-1-2 ——> 3-null-2 ——> null-3-2 ——> null-3-2 ——> 1-3-2 ——> 1-3-null ——> finished and yes
•  » » 5 years ago, # ^ |   0 Please notice the second condition among the three in the problem statement:2.pillar i contains exactly one diskAccording to this, you won't be able to perform your third (also fourth) move.
 » 5 years ago, # |   0 In problem B,what happens when there are two discs of same radius , doesn't it affect the answer? My both solutions taking the case of equal radii discs and ignoring that case are accepted. I'm little confuse here ,help me out.
•  » » 5 years ago, # ^ |   0 "The second line contains n integers a1, a2, ..., ai (1≤ai≤n), where ai is the radius of the disk initially placed on the i-th pillar. All numbers ai are distinct." — As the problem statement says.
•  » » » 4 years ago, # ^ |   0 Thanks for the beautiful explaination AQZZ, because a[i — 1] becomes a max and a[i] becomes a min. What does it mean?
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Shouldn't it be that: a[i] - end point  a[i-1]-starting point If not , then whats the reason? AQZZ
 » 3 years ago, # |   0 In Problem D's solution: Answer with value $best_i+sum(i-len,i-1)-k$.Is this a typo? I think it should be $best_i+sum(i+1,i+len)-k$.