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Today's atcoder Problem discussion
By sguu, history, 6 years ago, In English

Someone Please Explain D,E solution(English)?It's in Japanese

https://atcoder.jp/contests/aising2019

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6 years ago, # |
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My solution for D: It can be observed that the whole game consists of two phases: In the first phase Takahashi will take the greatest numbers, which Aoki will take some numbers around x. In the second phase the two will alternately take the greatest number. One should binary search on the position such where the transition between two phases happen. To check it one may need another binary search to count the numbers taken. After that the sum can be computed in O(1) if precalculated the partial sums and partial sum on even indices. The solution takes time.

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    6 years ago, # ^ |
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    Can u explain the two phase in above test case.Thanks

    5 5

    3 5 7 11 13

    1

    4

    9

    10

    13

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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it +32 Vote: I do not like it

      I'll explain the third query. I'll use T for Takahashi and A for Aoki. First T takes 13, A takes 7, and then T takes 11. Then A was supposed to take 11, but it is already taken by A. This is where the transition happens: the two players' territory(I don't know if this is clear enough) intersects. From now on, A and T will alternately take the greatest number remaining, which is: A takes 5 and T takes 3.

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    6 years ago, # ^ |
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    I had a similar idea: a suffix of A will be taken by Takashi, some part (same length or one less) before that by Aoki and the remaining prefix alternatingly. Then I processed the queries sorted decreasingly and kept pointers to the borders, so it's .

    edit: It's

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      6 years ago, # ^ |
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      Oh yes, this can be done more efficiently if processed offline. But I think that's due to the sorting process :D

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        6 years ago, # ^ |
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        Yeah, I added the in the wrong place :D

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    6 years ago, # ^ |
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    How to find the position of transition?

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      6 years ago, # ^ |
        Vote: I like it +28 Vote: I do not like it

      In fact, I binary searched on the suffix that was taken by Takahashi. which requires the length of the region of Takahashi to be at most one more than the length of the region of Aoki. This may be hard to explain, you may refer to my code. Code

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        6 years ago, # ^ |
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        why l=pos-1 in ur code?

        and ll numl=mid-(lower_bound(a,a+n+1,2*x-a[mid])-a) what it does?

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          6 years ago, # ^ |
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          The first one is because I choose r as the final answer, and pos is an valid answer. The second one implies the amount of numbers in Aoki's region.

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6 years ago, # |
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My solution for E: Let dpv, j, 0 denote the minimum sum of Ai in the connected component of v if we cut j edges in the subtree of v, and the connected component of v only consists of batteries. And dpv, j, 1 denote pretty much the same thing, but allows the connected component of v consists of computers. The DP should be calculated in an ordinary manner, which may seem to have a complexity of O(N3), but is actually O(N2). There are many proofs given on this kind of problems, e.g., Barricades in Algorithmic Engagements 2007, which can be found in the book looking for a challenge

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