### vovuh's blog

By vovuh, history, 22 months 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)

• +61

 » 22 months ago, # | ← Rev. 2 →   +6 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.
•  » » 22 months ago, # ^ |   +3 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.
 » 22 months ago, # |   0 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.
•  » » 22 months ago, # ^ | ← Rev. 3 →   0 Use a generator instead.P.S. My solution is plain simple O(N 3), 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.
•  » » » 22 months ago, # ^ |   0 I realized there is some people which use O(N^3) and can't pass the all star case. Thanks.
•  » » 22 months ago, # ^ |   +10 You can submit the code that generates test case of any size you want. See "generated input" tab in a hack window.
•  » » » 22 months ago, # ^ |   +2 Thanks.
•  » » 22 months ago, # ^ |   0 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
•  » » » 22 months ago, # ^ |   -9 why this contest is not rated yet??????
 » 22 months ago, # |   -8 why this contest is not rated yet??????
 » 22 months ago, # |   +1 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?
 » 22 months ago, # |   0 When will the rating be published ?
•  » » 22 months ago, # ^ |   0 maybe after system tests
•  » » » 22 months ago, # ^ |   0 why does is take so long? I mean there are only like 5 peoples in queue right now
•  » » » » 22 months ago, # ^ |   0 but when after judge over some solutions, another solutions will add in the queue
 » 22 months ago, # |   +2 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.
 » 22 months ago, # |   0 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
•  » » 22 months ago, # ^ |   0 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
•  » » » 22 months ago, # ^ |   0 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
 » 22 months ago, # |   +3 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!
•  » » 22 months ago, # ^ |   +3 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).
 » 22 months ago, # |   +5 I found another way to do D which is also really simple 41070174Did anyone else find another valid way?
•  » » 22 months ago, # ^ |   0 Actually, I did the same thing.
•  » » 22 months ago, # ^ | ← Rev. 2 →   +8 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.
 » 22 months ago, # |   +3 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.
 » 22 months ago, # |   +4 problem E — dp+difference array, that was a nice problem (Y)
 » 22 months ago, # |   +9 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.
•  » » 16 months ago, # ^ | ← Rev. 4 →   0 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[3] contains 6 and cols[6] 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
 » 22 months ago, # |   +3 Why does this soln gives a TLE https://codeforces.com/contest/1015/submission/41034740
•  » » 22 months ago, # ^ |   0
•  » » » 22 months ago, # ^ |   0 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.
 » 22 months ago, # |   0 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.
•  » » 22 months ago, # ^ |   +3 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.
•  » » » 22 months ago, # ^ |   +1 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?
•  » » » » 22 months ago, # ^ |   0 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
•  » » » » » 22 months ago, # ^ |   0 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.
•  » » » 22 months ago, # ^ |   0 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?
•  » » » » 22 months ago, # ^ |   0 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!
 » 22 months ago, # |   0 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?
•  » » 22 months ago, # ^ |   0 Yes
 » 22 months ago, # |   0 In problem E how the output is -1 for input 5 5 .*... **.. .... .*... .....Someone, please explain.
•  » » 22 months ago, # ^ |   0 Because you can't draw this grid with stars.
 » 22 months ago, # | ← Rev. 2 →   +3 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
 » 22 months ago, # | ← Rev. 4 →   0 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.
 » 22 months ago, # |   0 Why it 's wrong answer on test 26, i didn't know where is wrong? Please help me! Thanks! http://codeforces.com/contest/1015/submission/41127014
•  » » 22 months ago, # ^ |   0 It says it in the error message. You printed 1000000001, but there are only 1000000000 houses.
•  » » » 22 months ago, # ^ |   0 I printed the answer by using freopen and Ctrl+F (find) 1000000001 but not exist in my answer.
•  » » » 22 months ago, # ^ |   0 Thank you! I fixed my problem.
 » 22 months ago, # |   0 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.
•  » » 22 months ago, # ^ |   0 Yes, you are right. Thank you, fixed
 » 22 months ago, # |   0 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)
 » 22 months ago, # |   0 In problem D , what does the statement means "but you can't visit the same house in sequence"????
 » 22 months ago, # |   0 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 .
 » 22 months ago, # |   0 The std of E2 is TLE
 » 22 months ago, # | ← Rev. 3 →   0 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
 » 22 months ago, # | ← Rev. 2 →   0 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
•  » » 21 month(s) ago, # ^ |   0 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
 » 21 month(s) ago, # |   0 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
•  » » 21 month(s) ago, # ^ | ← Rev. 3 →   0 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.
 » 21 month(s) ago, # |   0 Can anyone explain the problem F with a example ? Thnaks !!
 » 21 month(s) ago, # |   0 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.
 » 19 months ago, # |   +5 Bad time limit of problem E2, my O(N^3) solution easily passed in TL.
 » 3 months ago, # |   0 Why is this code giving tle? My approach is same as the one mentioned in the tutorial!! https://codeforces.com/contest/1015/submission/71619845
•  » » 2 months ago, # ^ |   0 Mine also !! [https://codeforces.com/contest/1015/submission/74198722]Did you find the error?
 » 2 months ago, # | ← Rev. 2 →   0 can anyone explain logic behind prob a prefix sum method thanks
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2 months ago, # |
0

can anyone explain what this lines does in the code