### vovuh's blog

By vovuh, history, 4 years ago, translation, 1015A - Points in Segments

Tutorial
Solution (Vovuh, O(n + m))
Solution (Vovuh, O(n * m))

1015B - Obtaining the String

Tutorial
Solution (Vovuh)

1015C - Songs Compression

Tutorial
Solution (Vovuh)

1015D - Walking Between Houses

Tutorial
Solution (BledDest)

1015E1 - Stars Drawing (Easy Edition)

Tutorial
Solution (MikeMirzayanov, O(n^3))

1015E2 - Stars Drawing (Hard Edition)

Tutorial
Solution (Vovuh, O(n^2))

1015F - Bracket Substring

Tutorial
Solution (Vovuh) Tutorial of Codeforces Round #501 (Div. 3) 501, Comments (71)
 » 4 years ago, # | ← Rev. 2 →   Thanks for the editorial.I have one question about problem F dp recurrence, would the recurrence stated above be able to differentiate(seq)seqseqfromseq(seq)seqgiven seq is a valid bracket sequence, and the substring you are looking for.obviously in the two solutions the position, balance, and boolean parameter will be the same. Also the prefix of length K should be the same too.So I am asking, how will this dp handle situation such as this?UPD:Ok so after some thinking I realized that the dp state does not have to have a unique sequence, the only thing that has to be unique is the value it returns given a set of parameters. So seq(seq)seq and (seq)seqseq should be able to "reach the end" equal amount of ways, but I'm no so sure they can be made an equal amount of ways.
•  » » In my DP I do something special the first time the full pattern is encountered: I call another DP that doesn't track the pattern at all and simply counts the number of valid bracket sequences with remaining characters and balance number.
 » There is a problem that we can't hack the O(N^3) E2 solution. Because that our test cases is bigger than the limit->256 KB, but it can be a possible input.
•  » » 4 years ago, # ^ | ← Rev. 3 →   Use a generator instead.P.S. My solution is plain simple O(N3), and with N = M = 1000 test (all stars), it passes with 2430 ms. Fairly wide gap (for 3000 ms constraint), so I don't expect a lot of hacks with that test.
•  » » » I realized there is some people which use O(N^3) and can't pass the all star case. Thanks.
•  » » You can submit the code that generates test case of any size you want. See "generated input" tab in a hack window.
•  » » » Thanks.
•  » » maybe it's O(n^2 sigma{i,j} min(i,j,n-i,m-j))i write a program like this #include "bits/stdc++.h" using namespace std; typedef long long ll; int main() { int ans = 0; for(int i = 1 ; i <= 1000 ; ++ i) { for(int j = 1 ; j <= 1000 ; ++ j) { ans += min(min(i, j), min(1000 - i, 1000 - j)); } } printf("%d\n", ans); } it print "cnt = 166666500, time = 1.666665"and my program used 1.30s pass the all star without 4 corner stars data
•  » » » why this contest is not rated yet??????
 » why this contest is not rated yet??????
 » IN question E2 if it would have asked to find the minimum possible stars too and if many multiple answers exist print any. Then how could we solve it any idea anyone?
 » When will the rating be published ?
•  » » maybe after system tests
•  » » » why does is take so long? I mean there are only like 5 peoples in queue right now
•  » » » » but when after judge over some solutions, another solutions will add in the queue
 » FOR problem 'F' , can anyone explain me in simple words(possibly with an example) the dp strategy? I mean dp(i,j,k,l)part. Not able to understand tutorial.Also, not able to understand the recurrance part.
 » In Problem D editorial, I didn't understand the s - k + 1 in min(n - 1, s - k + 1). Can someone explain this please. thanks
•  » » So if s — remaining distance and k — remaining steps including current one, s - (k - 1) tells you how big the current step can be, so that after this step s >= (k-1), because you still need to make (k-1) of them and the smallest possible distance equals 1
•  » » » Oh! so basically, the size of the step I need to take so at least I can finish travelling the rest of the distance left using a step size of 1. Thanks
 » For the first problemCan someone help with these:How is the prefix sum calculation approach relevant to this problem (I fail to see) Why take size of cnt as m+2 Like why do --cnt[r+1] in the first for loop and why cnt[i]+=cnt[i-1] in the second for loopThanks!
•  » » This is the so-called "event scheduling" process. Basically each segment has a beginning and an ending index (l and r). If you were to add all points from l to r manually, you would get an O(nm) solution. However, notice that you can only add one to position l, and when you traverse the array once again, and use prefix sums, pref[i] will >= 1 if there is a point on position i. However, since there is no point in r+1, nor is there in any indices to come, we put -1 at that position so the prefix sum would become 0. Adding multiple segments allows the prefix sums to be equal to the number of segments covering point i at position pref[i]. Try some examples on paper for better understanding. Here is an example (l=1, r=2, l=0, r=5): The event schedule array looks like this: {1, 1, 0,  - 1, 0, 0,  - 1} And the pref array looks like this: {1, 2, 2, 1, 1, 1, 0} (the arrays are 0-indexed and pref[i] really shows the number of points on position i).
 » I found another way to do D which is also really simple 41070174Did anyone else find another valid way?
•  » » Actually, I did the same thing.
•  » » 4 years ago, # ^ | ← Rev. 2 →   Mine might be similar to yours, my distance was ceil(dist / k), and then I reduced my dist by that amount, and reduced k by 1. I can't prove why it's correct though.
 » For D, what I did was s = (s / k + 1) * (s % k) + (k — s % k) * (s / k) In other words took s / k + 1 steps back and forth s % k times and remaining steps were s / k long.
•  » » The same approach was done by me
 » problem E — dp+difference array, that was a nice problem (Y)
 » For E I have a shameful solution in O(nmlog(nm)). I keep prefix sums for how many stars I've seen after me, but I don't keep track of contiguous subsequences or anything. Instead, for each point (i, j) as middle of star, I binary search for the maximum length of the star centered there by using prefix sums and checking that sum in all four directions is at least my binary search value (it obviously cannot be more). Then I use segment trees to mark that a point is included in some star. I store all my stars in a vector, but before printing them out I query for each point on the grid that is * I check if it is involved in a star (either by row or column) using segment tree point max query (I am so eager to add log factors). If all the *'s belong to stars I print out all the stars I found in the binary searches. This passes the big data in 1715 ms.
•  » » 3 years ago, # ^ | ← Rev. 4 →   I also came to a solution, but did not use a segment tree. I just used a vector of sets.vector> rows(numRows);vector> cols(numCols);If the grid has a * at 3, 6 then rows contains 6 and cols contains 3.Similarly to the editorial, as I am inspecting a star, I have to mark a row of the grid as being covered by a ray of a star (and the same with the column). I use upper_bound on the above sets to find what points are covered by the rays, and remove those points.Since each point in rows and cols can be removed only once the total time to remove all the points is M*N*log(N) + N*Mlog(M) which is M*N*log(N*M) (removal from a set of size N takes log(N) time).And for each star we need to do two upper bound calls to find points to remove, so N*M*log(N*M) for that.Then, I intersect both vector of sets and if a point is in both, then it cannot have been covered by any ray.https://codeforces.com/contest/1015/submission/48593389
 » Why does this soln gives a TLE https://codeforces.com/contest/1015/submission/41034740
•  » »
•  » » » Ah thanks @test616.cpp, I understood . But my main concern is doing random shuffling during a online contest may take some extra time penalty and it's too often that most of the Java users use Arrays.sort() to sort an array and after getting a TLE verdict we may realise that we need to shuffle the array.
 » I am not able to understand the dp part in Problem F can anyone elaborate it more. I am also not able to understand the need of finding the first Len dp.
•  » » The main idea is if you obtain some prefix of s and trying to add some character which is not equal to the next character of s, your target is to obtain the maximum prefix of s with the last character you add. For example, if s = ((()(())((), you obtained the first six characters ((()(( and you have to add the character (, you want to get the prefix of s of maximum length (because you want to obtain the string s as soon as possible, right?), your string will be looks like ((()((( and the maximum prefix of s corresponding to the suffix of this string will be (((. The first dp is needed to make transitions in the second dp faster.And the second dp just calculates the number of prefixes of regular bracket sequences with the last k characters equals to the prefix of s of length k.
•  » » » As it is given that s must be a substring, then we can calculate how many ')' or '(' are needed to make it a regular string. For example in '( ) ) ) ( )' , we need two '(' to the left of s ,hence now we have to place 2 (4-2=2) more brackets(after balancing s) , those two can be(bold brackets) : ( ) ( ( ( ) ) ) ( ) or ( ( ) ( ( ) ) ) ( ) or ( ( ( ) ( ) ) ) ( ) or ( ( ( ) ) ) ( ) ( ) or ( ( ( ( ) ) ) ( ) ) so there are five places to put one pair in the string, now we can do that for any number of pairs, which is the catlan number, and we can put these in any gaps (except in s). Is this a correct approach?
•  » » » » Well... I don't sure you can solve this problem this way, because it is too hard to consider all valid placements of brackets for a random bracket string exactly once. Because of this we do dynamic programming
•  » » » » » for each gap we can have dp state such that if there are n gaps and x is the number of pairs that has been used then we can calculate dp[n][x] from dp[n-1][x],dp[n-1][x-1],dp[n-1][x-2]... and multiplying their respective catlan numbers . to speed up this.
•  » » » » tatya_bichu No it cannot be solved by this method, Initially i was also thinking on the same lines. But the problem is, we are trying to make the sequence before the given sequence regular and same for the sequence after the given sequence. I mean, if we create a sequence like seq1 given seq2 than seq1 will handle the closed brackets and be regular apart from this. and seq2 will handle open brackets and be regular apart from this. But It need not be the case.
•  » » » Thanks Vovuh. I understood the concept except the factor 'j' in dp(i,j,k,l).COuld you please explain what does 'balance' exactly mean?
•  » » » » In some sense the balance means the number of the opening brackets ( which are weren't closed on the current prefix. I.e. for empty prefix the balance is zero always. If you will add the opening bracket, the balance will increase by one. If you will add the closing bracket, the balance will decrease by 1. But there is one more thing. In any moment of time the balance should be greater than or equal to zero. It is because you should handle the case when the number of closing brackets is getting greater than the number of opening brackets. Btw you can try to understand it by yourself with this definition of balance. I hope it helps!
 » Problem E "If it is impossible to draw the given grid using stars only, print "-1"." So all dots case's answer is 0 instead of -1?
•  » » Yes
 » In problem E how the output is -1 for input 5 5 .*... **.. .... .*... .....Someone, please explain.
•  » » Because you can't draw this grid with stars.
 » 4 years ago, # | ← Rev. 2 →   I also have posted this in the Announcement:Hello, in problem F, why does this solution not work :read nread sringget delta of string : delta"(("=2 ; delta")"=-1 like if you have "(" add 1 and if you have ")"substract 1get the minimum over all deltas :code : int del=0;int mi=del;for(int j=0;j
 » 4 years ago, # | ← Rev. 4 →   Hi wanna ask how this code gets TLE? It is E2 solution with about lets say 10 nested loops which do about 10^6 operations ( each at most) , so how does not it pass in 3 seconds? Pretty clear code below: http://codeforces.com/contest/1015/submission/41120679 I'd be very thankful for any help.Edit: Fixed. Endl operation took super long.
•  » » It says it in the error message. You printed 1000000001, but there are only 1000000000 houses.
•  » » » I printed the answer by using freopen and Ctrl+F (find) 1000000001 but not exist in my answer.
•  » » » Thank you! I fixed my problem.
 » In the F tutorial: which equals to the suffix of the prefix of s of length lenI think this len should be i?Sorry for poor English.
•  » » Yes, you are right. Thank you, fixed
 » Can someone explain me what it means: "wrong answer 1 <= x_99 — s_99 should be satisfied" in E1 on test No20? Here is my solution (I know that this is a bad code, but I did my best)) http://codeforces.com/contest/1015/submission/41223384)
 » In problem D , what does the statement means "but you can't visit the same house in sequence"????
 » In 1015C (Songs Compression) , if I want to compress the i'th song why Sum will decreased by a[i] — b[i] in every iteration ?? " where Sum = summation of all a[i] " . I understand the whole tutorial except this point . It will be very helpful for me if anybody clarify it . Thank you very much .
 » The std of E2 is TLE
 » 4 years ago, # | ← Rev. 3 →   Is it legal to get TL with String.format("%d %d %d") and pass tests with print(i1);print(i2);... ? Is String.format slow? Is java slow? Am I missing something?http://codeforces.com/contest/1015/submission/41540968http://codeforces.com/contest/1015/submission/41540991
 » 4 years ago, # | ← Rev. 2 →   In Stars Drawing (Hard Edition) Please explain why? Let's increase h[i,j-len] by one and decrease h[i,j+len] by one. Thanks
•  » » We use the principle of prefix sums:if (i > 0) v[i][j] += v[i - 1][j];if (j > 0) h[i][j] += h[i][j - 1];If there was a positive number at the point (i,j), then it affects all the points that stand after it until it meets the negative number. For example, you can read about it here
 » I am not getting the editorial for F.can somone explain me What is len[i][j]Plzz explain!!Let leni,j will denote the maximum length of the prefix of s which equals to the suffix of the prefix of s of length i with the additional character '(' if j=0 and ')' otherwise. In other words, leni,j
•  » » 4 years ago, # ^ | ← Rev. 3 →   Do you know how does Knuth-Morris-Pratt algorithm work (especially "Partial match" table) and why does it work? If not, start with that. It might not be easy, I know, but there are multiple resource on this in the Internet and I won't be able to explain it better than it was already done in some articles / videos. Afterwards this information should be pretty straightforward as it is the same concept.
 » Can anyone explain the problem F with a example ? Thnaks !!
 » In problem F you can eliminate the flag l by simply not changing the value of k once k hits |s|, this is how you would construct a Deterministic Finite Automaton which checks whether some string s appears at least once in the input string — you simply make the outgoing edges at vertex |s| go back to |s|. The code is also somewhat simpler.
 » Bad time limit of problem E2, my O(N^3) solution easily passed in TL.
 » Why is this code giving tle? My approach is same as the one mentioned in the tutorial!! https://codeforces.com/contest/1015/submission/71619845
•  » » Mine also !! [https://codeforces.com/contest/1015/submission/74198722]Did you find the error?
 » 2 years ago, # | ← Rev. 2 →   can anyone explain logic behind prob a prefix sum method thanks
»

can anyone explain what this lines does in the code

# endif

•  » » It is for getting input from input.txt, which is a file that should be present in your current directory. You can paste the input there and your program will read input from this file. Same goes for the output.txt.
 » I solved problem D using binary search. Here's what I did:1) Ofcourse if k>s or k*(n−1) 88882383
 » Help... Why this F solution is wrong : https://codeforces.com/contest/1015/submission/95085826Idea : Let's say our final string would be A where at least 1 substring of S is present. Fix the starting position of the substring S, X = current starting position of S, Y = current end of S. In our dp, if pos < X || pos > y, that means we can choose this position to be either "(" or ")", so we call function for both of them. When pos >= X && pos <= Y, we don't have a choice, we have fixed that length for S, so we call function according to what we have in S. For each possible X (starting position can be 0,1,2...) do a new dp, and increase our answer.