Codeforces Round #359 (Div. 2)D. Kay and Snowflake【树的重心】

1. 求某个结点子树的重心
2. 情况1：重链上的儿子结点i为根的树结点数小于父结点j为根的树节点数，j为重心
3. 情况2：与1相反时，重心必在j到i为跟的树的重心的路径上

/* ***********************************************Author        :MaltubEmail         :[e-mail]          :htttp://www.xiang578.com************************************************ */#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>//#include <bits/stdc++.h>#define rep(i,a,n) for(int i=a;i<n;i++)#define per(i,a,n) for(int i=n-1;i>=a;i--)#define pb push_backusing namespace std;typedef vector<int> VI;typedef long long ll;const ll mod=1000000007;const int N=300000+10;vector<int>g[N];int n,q,son[N],fa[N],ans[N];void dfs(int x){    son[x]=1;    ans[x]=x;    for(int i=0; i<g[x].size(); i++)    {        int y=g[x][i];        dfs(y);        son[x]+=son[y];    }    for(int i=0;i<g[x].size();i++)    {        if(son[g[x][i]]*2>son[x])            ans[x]=ans[g[x][i]];    }    while(2*(son[x]-son[ans[x]])>son[x])    {        ans[x]=fa[ans[x]];    }}int main(){    int t;    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    scanf("%d%d",&n,&q);    for(int i=1; i<=n; i++)        g[i].clear();    for(int i=2; i<=n; i++)    {        scanf("%d",&fa[i]);        g[fa[i]].push_back(i);    }    dfs(1);    for(int i=0; i<q; i++)    {        scanf("%d",&t);        printf("%d\n",ans[t]);    }    return 0;}