### chokudai's blog

By chokudai, history, 5 months ago, We will hold M-SOLUTIONS Programming Contest 2021(AtCoder Beginner Contest 232).

The point values will be 100-200-300-400-500-500-600-600.

We are looking forward to your participation! Comments (34)
 » C++20 support when
 » 5 months ago, # | ← Rev. 2 →   How to solve C ? I'm just getting WA on one case and AC on the other 21 — Submission sad noises
•  » » 5 months ago, # ^ | ← Rev. 2 →   lmao I did the same exact thing, I would like to know why this doesn't work as well
•  » » » 5 months ago, # ^ | ← Rev. 2 →   Yes even i did the same thing.. and got WA. I realized later that "the degree sequence of two graphs should be same to be isomorphic" — this is a necessary but not sufficient condition.So what I did was to store the component sizes of each node of two graphs and checked if both are equal. Luckily it passed. But this too is wrong!I guess the test cases are weak.This the the ac code which I wrote: However this code gives wrong answer on this test case: 6 5 1 2 2 3 3 4 4 5 3 6 1 2 2 3 3 4 4 5 2 6 The answer should be "No" but my code prints "Yes"I didn't see the constraints initially (stupid mistake I know). But the intended solution is to use brute force
•  » » Your solution is actually very wrong, it should definitely not pass that many testcases. Graph isomorphism is a hard problem, in here you just need to enumerate all permutations P from the statement and check whether the statement condition holds. You can do that with std::next_permutation().
•  » » » » I also did the same way and got AC but my condition to check whether the permutation is valid or not is different.Here is my Submission: https://atcoder.jp/contests/abc232/submissions/28008891
•  » » » » I think the problem might be usage of a set. It will discard equal elements. Use multiset instead.
•  » » » My submission also fails on only one test case. Weak tests I guess. next_permutation worked.
•  » » Use next_permutation()
•  » » 5 months ago, # ^ | ← Rev. 5 →   Use the edge connectivities, not the degrees. Two graphs might have the same degrees but aren't isomorphic.Try this input 6 5 1 2 2 3 2 4 4 5 5 6 1 2 2 3 3 4 2 5 5 6 
•  » » » Can you please share an example of when the degrees are equivalent but the graphs are not isomorphic?
•  » » » » _\_ _\_ and _\_\_ _
•  » » » » I added an example to my comment.
•  » » » » Check the structures of 2-Methylpentane and 3-Methylpentane. Same degrees, but not graph isomers.
•  » » Brute force all the permutations using std::next_permutation() and check if the condition mentioned holds true
 » can we solve F with flow.
•  » » I tried but somehow it got WA, I still don't get why.
 » For problem E, how do you find theses DP optimization ideas ?
 » Can anyone explain, how we are getting 4 distinct value from f(k,i,j) and how we are forming those transitions? It would be a great help. Thanks!
•  » » "for each k, there are at most 4 distinct values of f(k,i,j)" : case 0 (i=0, j=0): Current row != Dest row. Current column != Dest column case 1 (i=0, j=1): Current row != Dest row. Current column == Dest column case 2 (i=1, j=0): Current row == Dest row. Current column != Dest column case 3 (i=1, j=1): Current row == Dest row. Current column == Dest column My submission uses 4x4 = 16 transitions, although some transitions are impossible and can be simplified.
 » 5 months ago, # | ← Rev. 7 →   My approach for E is different from the editorial, and I'm getting wrong answer on a few test cases. Any help would be much appreciated. Submission: LinkApproach: We are at (x1,y1) and want to reach (x2,y2) after exactly k operations. in one operation we can change either x or y. Let's say that we change x in p operations and y in k-p operations, and let dp1[p] be the number of ways to start from x1 and end at x2 after exactly p operations and dp2[k-p] be the number of ways to reach from y1 to y2 after exactly p-k operations, then I will add C(k,p)*dp1[p]*dp2[k-p] to my answer. (C(k,p) is the number of ways to select p operations that will change x out of the k operations)Now as far as dp1 and dp2 are concerned. We can precalculate both of then as follows: dp1[i] = (w-1)^(i-1) — dp1[i-1]dp1 is 1 if x1 = x2 and 0 otherwise dp2[i] = (h-1)^(i-1) — dp2[i-1]dp2 is 1 if y1 = y2 and 0 otherwiseCan anyone point out the flaw in this approach?Update: nevermind I interchanged h and w.
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5 months ago, # ^ |
Rev. 2

How about this $(3 \times 2)$ chessboard?

E E
Source E
Target E

There is exactly one way to go from source to target in 2 moves, while your code reports it to be zero.

•  » » The approach is fine. I implemented the same method and got AC. Here is my implementation. Sub
 » 5 months ago, # | ← Rev. 2 →   One of the best atcoder contests
 » Can someone explain the solution of problem E.
•  » » 5 months ago, # ^ | ← Rev. 3 →   I solved it using recursive DP. We just have to maintain that the if our current row matches with the final row or not, and our current column matches with current column or not( Actual row and column values don't matter here ,we just need to maintain that they are equal to final ones or not, so we can use 0/1 to denote that) . So currently if we are in same row as the final row, we can go to any other row and it will be equivalent (n-1 such rows) . If we are in some other row, we have two choices, either we can go to the final row or some other row (n-2 such rows). So similarly we can do it for columns. After K moves we can check if our both row and column are equal to final row and column or not. dp will look like dp[K] and each transition will be done in constant time . So overall Time complexity will be O(K).
•  » » » 5 months ago, # ^ | ← Rev. 3 →   thanks for your hint, I solved this problem. the interesting thing is this code will TLE on my computer, but got AC after submit it. codell dfs(int k, int r, int c) { if (K == k) return r && c; ll& ans = dp[k][r][c]; if (ans != -1) return ans; ans = 0; if (r && c) { ans = dfs(k + 1, 0, 1) * (n - 1) % mod + dfs(k + 1, 1, 0) * (m - 1) % mod; } if (r && !c) { ans = dfs(k + 1, 0, 0) * (n - 1) % mod + dfs(k + 1, 1, 0) * (m - 2) % mod + dfs(k + 1, 1, 1); } if (!r && c) { ans = dfs(k + 1, 1, 1) + dfs(k + 1, 0, 1) * (n - 2) % mod + dfs(k + 1, 0, 0) * (m - 1) % mod; } if (!r && !c) { ans = dfs(k + 1, 0, 0) * (n - 2 + m - 2) % mod + dfs(k + 1, 1, 0) + dfs(k + 1, 0, 1); } ans %= mod; return ans; } update: it's beacause of stack size, it works after i add --stack=536870912 to compiler flag.
•  » » » How do you solve problem F?
 » I did problem D using BFS but getting TLE.Anyone know how to resolve?Problem link:-Problem-D
•  » » I also solved the mentioned problem using BFS but got no TLE. Could you share your submission?
•  » » » My Submission-HERE
•  » » » » Your function fun is very bad in more than one way (why recursion? why undirected graph when we are only allowed to move right and down?) and can't handle a 100x100 grid filled with '.' because it becomes too slow. You can change it to something like this: Spoilervoid fun(vector &v, vector> &adj, ll H, ll W) { for (int col = 0; col < W; col++) { for (int row = 0; row < H; row++) { if (v[row][col] != '.') continue; if (col + 1 < W && v[row][col + 1] == '.') adj[row * W + col].pb(row * W + (col + 1)); if (row + 1 < H && v[row + 1][col] == '.') adj[row * W + col].pb((row + 1) * W + col); } } } But even building an adjacency list is an overkill.
•  » » BFS should work just fine for this problem (example). But be sure not to visit the same cell more than once. If this isn't done, then a 100x100 grid filled with '.' would fail with TLE.