[CF592D] Super M - 树的直径

Description

有一个有 n 个节点的树，其中有 m 个点被标记。现在从任意点开始，去遍历这 m 个点，问你遍历这 m个点的最少路程（边的权值为1）。

Solution

答案显然是这个标记点构成的虚树大小减去虚树直径

当然我们不需要真的建虚树，从一个标记点开始 DFS，考察子树内是否有标记点即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

int n, m;

const int N = 1e6 + 5;

vector<int> g[N];
int dep[N], siz[N], tag[N];

void dfs(int p, int from)
{
    siz[p] = 0;
    for (int q : g[p])
    {
        if (q == from)
            continue;
        dep[q] = dep[p] + 1;
        dfs(q, p);
        siz[p] += siz[q];
    }
    if (tag[p] || siz[p])
        ++siz[p];
}

signed main()
{
    ios::sync_with_stdio(false);

    cin >> n >> m;
    for (int i = 1; i < n; i++)
    {
        int x, y;
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }

    int s = 1e9;
    for (int i = 1; i <= m; i++)
    {
        int x;
        cin >> x;
        s = min(s, x);
        tag[x] = 1;
    }

    if (m == 1)
    {
        cout << s << endl
             << 0 << endl;
    }
    else
    {
        dep[s] = 1;
        dfs(s, 0);
        for (int i = 1; i <= n; i++)
            dep[i] *= tag[i];
        int a = 0;
        for (int i = 1; i <= n; i++)
            if (dep[i] > dep[a])
                a = i;
        dep[a] = 1;
        dfs(a, 0);
        for (int i = 1; i <= n; i++)
            dep[i] *= tag[i];
        int len = *max_element(dep + 1, dep + n + 1);
        int b = 0;
        for (int i = 1; i <= n; i++)
            if (dep[i] > dep[b])
                b = i;
        cout << min(a, b) << endl
             << 2 * (siz[a] - 1) - len + 1 << endl;
    }
}