$f[i][0]$：第i个结点被父节点观察
$f[i][1]$：第i个结点被至少一个子节点观察
$f[i][2]$：第i个结点本身有哨兵
$f[i][0] = ∑min(f[j][1], f[j][2])$
$f[i][1] = 枚举k有哨兵∑(f[k][2] + f[i][0] - min(f[k][1], f[k][2]))$
$f[i][2] = ∑min(f[j][0], f[j][1], f[j][2])$

#include<bits/stdc++.h>
using namespace std;
const int N = 1505;
int h[N], ne[N], w[N], e[N], idx, not_root[N];
int f[N][3];
void add(int a, int b) {
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}
void dfs(int node) {
    f[node][2] = w[node];
    for (int i = h[node]; i != -1; i = ne[i]) {
        int j = e[i];
        dfs(j);
        f[node][0] += min(f[j][1], f[j][2]);
        f[node][2] += min(f[j][0], min(f[j][1], f[j][2]));
    }
    f[node][1] = 1e9;
    for (int i = h[node]; i != -1; i = ne[i]) {
        int j = e[i];
        f[node][1] = min(f[node][1], f[j][2] + f[node][0] - min(f[j][1], f[j][2]));
    }
}
int main()
{
    memset(h, -1, sizeof(h));
    int n;
    scanf("%d", &n);
    for(int i = 0;i < n;i++) {
        int id, m, wg;
        scanf("%d%d%d", &id, &wg, &m);
        w[id] = wg;
        for(int j = 0;j < m;j++) {
            int b;
            scanf("%d", &b);
            add(id, b);
            not_root[b] = 1;
        }
    }
    int root;
    for(int i = 1;i <= n;i++) {
        if(!not_root[i]) root = i;
    }
    dfs(root);
    printf("%d", min(f[root][1], f[root][2]));
    return 0;
}