A new city has just been built. There're  interesting places numbered by positive numbers from  to .

In order to save resources, only exactly  roads are built to connect these  interesting places. Each road connects two places and it takes 1 second to travel between the endpoints of any road.

There is one person in each of the places numbered  and they've decided to meet at one place to have a meal. They wonder what's the minimal time needed for them to meet in such a place. (The time required is the maximum time for each person to get to that place.)

题意：给你一个n个节点n-1条边的树，其中k个人在某些节点上，他们希望到同一个地方的时间最小（时间是最后一个人到的时间）。

思路：其实很简单，只需要求出来包含这么多节点的最小树的直径向上取整就行了，当时一直往重心上想。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
vector<int> e[maxn];
map<int, int> mp;
int cnt, ans;
int l, k;
 
void dfs(int u, int fa, int len)
{
    if(cnt > k)
        return;
    if(mp[u])
    {
        ++cnt;
        if(len > ans)
        {
            ans = len;
            l = u;
        }
    }
    for(int i = 0; i < e[u].size(); ++i)
    {
        int v = e[u][i];
        if(v == fa)
            continue;
        dfs(v, u, len + 1);
    }
}
 
int main()
{
    int n;
    scanf("%d%d", &n, &k);
    for(int i = 1; i < n; ++i)
    {
        int u, v;
        scanf("%d%d", &u, &v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    int tmp;
    for(int i = 1; i <= k; ++i)
    {
        int u;
        scanf("%d", &u);
        mp[u] = 1;
        tmp = u;
    }
    cnt = ans = 0;
    dfs(tmp, 0, 0);
    cnt = 0;
    dfs(l, 0, 0);
    printf("%d\n", (ans + 1)/2);
    return 0;
}