CF685B-求有根树每颗子树的重心

题目大意:

标题说的很清楚了.

题目思路:

根据树的重心的性质:两颗子树合并，新树重心一定在两个子树重心的连线上.

所以合并子树的时候，让子树们的重心往祖先跳。直到跳到重心结束.
PS:判定重心: 2 * 最大子树sz <= sz[u]即可。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define vll vector<ll>
#define fi first
#define se second
const int maxn = 3e5 + 5;
const int mod = 1e9 + 7;
int sz[maxn] , res[maxn] , f[maxn] , mx[maxn];
vi e[maxn];
void dfs (int u , int fa){
    
    sz[u] = 1; res[u] = u;f[u] = fa;
    for (auto & v : e[u]){
    
        if (v == fa) continue;
        dfs(v , u);
        sz[u] += sz[v];
        mx[u] = max(mx[u] , sz[v]);
    }
    for (auto & v : e[u]){
    
        if (v == fa) continue;
        int now = res[v];
        // 爆跳
        bool ok = false;
        do{
    
            int tmx = max(mx[now] , sz[u] - sz[now]);
            if (2 * tmx <= sz[u]){
    
                res[u] = now;
                ok = true;
                break;
            }
            now = f[now];
            if (ok) break;
        }while (now != u);
    }
    return ;
}
int main()
{
    
    ios::sync_with_stdio(false);
    int n , m; cin >> n >> m;
    for (int i = 2 ; i <= n ; i++) {
    
        int x; cin >> x;
        e[x].pb(i);
    }
    dfs(1 , 0);
    for (int i = 1 ; i <= m ; i++){
    
        int x; cin >> x;
        cout << res[x] << endl;
    }
    return 0;
}