iLoveIOI's blog

Atcoder Beginner Contest 257 Discussion?
By iLoveIOI, history, 2 years ago, In English

I couldn't find a discussion blog for this contest so here's one.

Does anyone know how to solve Ex?

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2 years ago, # |
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For problem D, my solution with Binary Search + BFS works fine except for 01_random_03.txt. Any hint ?

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    2 years ago, # ^ |
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    Did you binary search till 4e9 becauese I was facing same verdict till i realized cost could be 4e9.

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      2 years ago, # ^ |
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      My solution also fails on the same test case. Is there anyone who could look into my code.

      code

      I have used #define int long long

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        2 years ago, # ^ |
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        Overflow. P can be 10^9 10^11 * 10^9 = 10^20 which is too big even for long long.

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      2 years ago, # ^ |
        Vote: I like it -6 Vote: I do not like it

      explain your solution

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    2 years ago, # ^ |
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    With 4e9 it works fine. thanks.

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2 years ago, # |
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Submission In problem G why picking the longest transition (by prefix function) works?

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    2 years ago, # ^ |
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    Greedily removing the longest suffix of T that is a prefix of S seems to work as well.

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      2 years ago, # ^ |
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      Yeah did so after the contest and it passed but I am unable to prove it why does it work.

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        2 years ago, # ^ |
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        Because ans[i]<=ans[j] for i<=j because you can always construct a solution for index i from the constructive words of solution j by either deleting some characters of the last word (giving the same ans) or deleting some words then some characters (giving smaller ans).

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2 years ago, # |
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D was a nice variation of Floyd Warshall algorithm

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    2 years ago, # ^ |
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    BFS is a nice variation of Floyd Warshall?

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      2 years ago, # ^ |
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      You can also solve it using a slight modification of floyd warshall by defining

      $$$ w[i][j] = \frac{|x_i-x_j|+|y_i-y_j|}{p[i]}$$$

      and changing + operation with max

      code

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    2 years ago, # ^ |
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    I used this idea as well, but I think that binary search solution is also very nice.

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2 years ago, # |
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I have failed on F because of a little mistake in my BFS :)

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2 years ago, # |
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Problem C? Can anyone mention the approach? Thanks!

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    2 years ago, # ^ |
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    We can multiply all w[i]=w[i]*2, then check all w[i]+1 and w[i]-1 to be the possible best answer.

    Each check can be done in O(logn) by sorting childs and adulds separate, then binary search the number of correct guessed ones for fixed x.

    Edit: Or use prefix sums. Atfer sorting w[], we can split w[] at any position, then the correct guessed number is the number of childs left of the split position plus the number of adults right of the split position.

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    2 years ago, # ^ |
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    Keep track of score0 and score1. Use map to iterate the weights and move the border line of guessing 0 vs guessing 1 from left to right. I submitted a simplified version of the code at: https://atcoder.jp/contests/abc257/submissions/32757902

    I was unlucky to only solved it one minute after the contest ended.

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    2 years ago, # ^ |
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    check for all w[i], then also check for smallest weight-1 and largest weight+1(Solution)

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    2 years ago, # ^ |
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    just sort the children array and parent array and apply binary to get the ans. also check if we have only children or only adults

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2 years ago, # |
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How to solve E?

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2 years ago, # |
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can any one help me with D , I think I miss understood the question . They are saying that : "Takahashi's objective is to become able to choose a starting trampoline such that he can reach any trampoline from the chosen one with some jumps."

some jumps means we can assume that he will be able to jump for each consecutive pair right ?

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2 years ago, # |
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Hi Everyone, Can anyone help to find the mistake in my code of F? Link to Solution

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2 years ago, # |
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For problem F, my solution is giving runtime error on 27 hidden cases. Why? https://atcoder.jp/contests/abc257/submissions/32791294

UPD: I have done the same thing as editorial except DFS instead of BFS, which I don't think should make any difference

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    2 years ago, # ^ |
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    It makes all the difference, because the graph isn't necessarily a tree/forest, so there may be more than one simple path between nodes of different lengths, and DFS doesn't necessarily find the shortest of them

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