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  • Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round) D. Peculiar apple-tree

    D. Peculiar apple-tree
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n. Inflorescence number 1 is situated near base of tree and any other inflorescence with number i (i > 1) is situated at the top of branch, which bottom is pi-th inflorescence and pi < i.

    Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a-th inflorescence gets to pa-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time.

    Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.

    Input

    First line of input contains single integer number n (2 ≤ n ≤ 100 000)  — number of inflorescences.

    Second line of input contains sequence of n - 1 integer numbers p2, p3, ..., pn (1 ≤ pi < i), where pi is number of inflorescence into which the apple from i-th inflorescence rolls down.

    Output

    Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.

    Examples
    Input
    Copy
    3
    1 1
    Output
    1
    Input
    Copy
    5
    1 2 2 2
    Output
    3
    Input
    Copy
    18
    1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4
    Output
    4
    Note

    In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them.

    In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.

    分析:

    因为只有同一时间在同一地点的苹果相碰会消失,那么相碰一定只出现在同一层的苹果之间,且在该层苹果大于1的时候,一定会出现相碰。

    所以我们可以将苹果分层进行讨论

    如果该层的苹果数为奇数,那么无论经过怎样的过程,到最后始终会剩一个。

    如果为偶数,那么到最后一定一个也不剩。

    将每层得到的结果累加即可

    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    typedef long long LL;
    vector<int>V[100010];
    int num[100010];
    int maxx;
    void dfs(int id,int k)
    {
      num[k]++;
      maxx=max(k,maxx);
      for(int i=0;i<V[id].size();i++)
       dfs(V[id][i],k+1);
    }
    int main()
    {
        int n,x,ans;
        cin>>n;
        maxx=0;
        ans=0;
        for(int i=2;i<=n;i++)
        {
            cin>>x;
            V[x].push_back(i);
        }
        dfs(1,1);
        for(int i=1;i<=maxx;i++)
        {
          if(num[i]%2==1)
          ans++;
        }
        cout<<ans<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/a249189046/p/8513565.html
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