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### DarthKnight's blog

By DarthKnight, history, 2 years ago, , Writer: DarthPrince

Writer: DarthPrince

Writer: DarthPrince

Writer: DarthPrince

Writer: DarthPrince

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Writer: DarthPrince Tutorial of Codeforces Round #459 (Div. 1) Tutorial of Codeforces Round #459 (Div. 2) 459, Comments (53)
 » Bonus question for Div1B: try to solve when sigma can be of size of m.
•  » » Could you please explain how to solve this (bonus) problem? I've no idea...
•  » » » Probably something likeDP[u][v] is the maximum character needed for a person to be on u (on his turn), the other on v and the first wins no matter what.
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   Yes ,you are right.I misunderstood it. Expected Complexity:O(n*n*n) SpoilerMy idea is to maintain a dp state having [1st player position],[2nd player position],[edge number in adjacency list of 1st player from which he can take an edge].(Here assuming adjacency lists are sorted by symbols on edge). Now dp[i][j][k]=dp[i][j][k+1] dp[i][j][k]=max(dp[i][j][k],dp[j][x][y]). where x is kth element in adjacency list of i and y is the lowest position in adjacency list of j which contains a symbol greater than or equal to that present on edge i to x(which can be found using binary search) or do a two pointers.Complexity:O(n*n*n) Accepted code
•  » » » » » O(n^3 * logn)? The solution I described is O(n * (n + m)). Here's a submission of that solution: http://codeforces.com/contest/917/submission/34697087
•  » » » » » » Ohh I am sorry.I was thinking something different.Yes you are right!!!
•  » » » » » » How is this O(n(n+m)) ?
•  » » » » » » » Every (u, v) edge will be considered in state [u][i] for any i, so the transition is O(n*m) and there are O(n*n) states.
•  » » » » » » » » There are n*n states and each transition is n*m so isnt the complexity O(n*n*(n+m)) instead of O(n*(n+m))
•  » » » » » » » » » 2 years ago, # ^ | ← Rev. 2 →   No, because if you consider each edge, it will be considered only on the states that the first state is the vertex it starts. So each edge is considered for n states (the first state is fixed but the second can be any of the n).
•  » » teja trying to be smart but fails :P
•  » » Congrats, you just caused this problem to become of positive value :P
 » 2 years ago, # | ← Rev. 2 →   I have a (maybe simpler and elegant?) solution for div1 A.How would you solve this if no question marks were involved? Well, this is a standard problem. You iterate through the possible substrings and keep a score which starts from 0, which represents how many open brackets we have. Whenever you iterate over a '(', you add 1 and when you iterate over ')', you subtract one. The string is valid iff at the end the score is 0 and at no point during the iteration was the score negative.For the problem at hand, we do the exact same thing, except we keep one more counter, say qmarks, which represents the number of question marks so far. Obviously, when we iterate over '?'. we increment this counter. Also, if at any point of the iteration we have qmarks > score, then at least one of the question marks has to be an open bracket (if they were all closed brackets, we would have negative score, which is illegal). Therefore, if this situation occurs, increment score and decrement qmarks.At the end, a substring is legal iff its length is even and qmarks >= score (we can use question marks to close off all remaining open brackets).Seems simpler than the solution in the editorial (and also doesn't use additional memory, if that's relevant).
•  » » simple solution, thank u so much...u saved me from official solution
•  » » Nice One :) .
•  » » Thanks for such a nice solution.
•  » » Huge thanks!!!!!
•  » » I had basically the same solution but (again maybe?) just a little more natural to explain.At each moment we will define l and r as minimum and maximum number of unclosed brackets in the sequence before given that it is a prefix of a correct bracket sequence. For every starting position initially l = r = 0. When we iterate over '(' or ')' we obviously increase or decrease both by one correspondingly. When we iterate over '?' we decrease l and increase r by one. If r < 0 at any point we stop here and go to a next starting position. l can't be less than 0, so if l < 0 we increase it by 2. Finally the ending position is possible if and only if l = 0. Codestring s; getline(cin, s); int n = s.size(); int ans = 0; forn(st,0,n) { int l = 0; int r = 0; forn(e,st,n) { if (s[e] == '(') { l++; r++; } else if (s[e] == ')') { l--; r--; } else if (s[e] == '?') { l--; r++; } if (r < 0) break; if (l < 0) l+=2; if (l == 0) ans++; } } cout<
•  » » Could you please explain it on a test: "???(" ?. Here, if I consider the whole substring, right from the 1st iteration, qmarks > score. so: qmarks = 1 score = -1; qmarks = 2; score = -2; qmarks = 3; score = -3; score = -2Hence, the length is even and qmarks >= score. Hence it's legal. But you can't form any valid bracket with this sequence. Am i missing something?
•  » » » When qmarks > score, you decrement qmarks and increase score, not the other way around. The logic is that you turn a question mark into an open bracket, since at least one of the question marks has to be open.
•  » » » » Thanks a lot. I must read carefully!
•  » » Thank you so much. I don't understand why editorialists don't write editorials like this.
•  » » Won't the time complexity be O(n^3) then?O(n^2) for every every substring and checking takes another O(n).?
 » Solution #2 for problem D is just beautiful!
•  » » And India is blessed to have a coder like you.
 » Div 1 E : "First two variables can be calculated using aho-corasick and segment tree." Can someone explain how to do this? Thanks in advance.
 » How can i solve div2 C with dp?
 » I hope someone can answer my confusion...To Div2 D,I still can't understand the rightness of the editorial.Can this ensure that both players play optimally?
•  » » 2 years ago, # ^ | ← Rev. 2 →   if one's make "a" move can make other lose then it is optimally move right? so you have to find a path that the ichar(i) is big enough that the next person cant go any furtherc < int(ch(v, x) - 'a') and dp(u, x, ch(v, x)) = false its mean the ichar you have is larger than the limit now and the ichar you have ,can make the next player fail to move,( dp is false) , thus the state now is true
•  » » » Thanks a lot!
•  » » » I steal couldn't understand the problem solution :(
•  » » » » maybe you shoulde study classical sg function first
•  » » » » » what is " sg function " ?
•  » » » » » » Sprague-Grundy theorem you can google it
•  » » » » » » » How Sprague-Grundy theorem can be used here?? can you explain?? It seems that only DP is used here.
•  » » » » » » » » I made a mistake...sorry..
 » Could anyone please tell me what is sigma in 917B editorial?
•  » » 2 years ago, # ^ | ← Rev. 3 →   Sigma is the size of the constraint on the edge. In this case,its 26. If we allowed numbers(from 0 to 9) and alphabets both in the problem, then it would have been 36.
•  » » » thanks
 » Could anyone confirm that this solution is actually O(N^4)? I have a nested loop in the recurrance function which already is O(N^2).
 » Can someone please explain more precise how to make the the dp with matrix multiplication on Div1C/Div2E? Especially how to construct the matrix to be multiplicated and the initial step of the dp? I have tried to mulptiply the matrix m[i][j] as cost to go from state number i to state number j, but I couldn't find the recurrence
 » 2 years ago, # | ← Rev. 4 →   admin, problem E please add this test,it can make my AC code run out of 3 second:(4.4s)probably this isn't the worst test of my programmethe upper_bound complexity of my programme is O(n*log(n)^2*sqrt(Q)) if we cha carefully it can be failed system test.... n=1e5; m=400; num_q=1e5; int i,j,s,p,q,u,v; char ch; for(i=0;i<=100000;i++) { if(i==0) ex[i]=1; else ex[i]=ex[i-1]*base%mod; } for(i=0;i