【掩体计划——DFS+缩点】

题目

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
vector<vector<int>> g;
bool st[N];
int ans = 1e9;
bool dfs(int f, int u, int dis)
{
    bool is = 1;
    for (auto j : g[u])
    {
        if (j == f)
            continue;
        is &= dfs(u, j, dis + (g[u].size() > 2));
    }

    if (g[u].size() == 1) // 无向图叶子节点判定
        is = st[u];
    if (is)
        ans = min(ans, dis);
    return is;
}
int main()
{
    int n;
    cin >> n;

    g.resize(n + 1);
    for (int i = 1; i < n; i++)
    {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    g[1].push_back(-1); // 为了统一那个多出来的父节点
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++)
    {
        int x;
        cin >> x;
        st[x] = 1;
    }

    dfs(-1, 1, 0);

    cout << ans;
    return 0;
}