十分简单的一道题，令我脑部旋转

#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, m, t, ans, a[N], vis[N], ind[N];
vector <int> p[N];
void pus(int x) {
	vis[x] = 1;
	for (int i = 0; i < p[x].size(); ++i) {
		int to = p[x][i];
		if (vis[to]) continue;
		pus(to);
	}
}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &t);
		for (int j = 1, v; j <= t; ++j)
			scanf("%d", &v), p[v].push_back(i);
	}
	scanf("%d", &m);
	for (int i = 1; i <= m; ++i) {
		memset(vis, 0, sizeof(vis));
		for (int j = 1; j <= n; ++j) scanf("%d", &a[j]);
		for (int j = 1; j <= n; ++j) 
			if (!a[j])
				pus(j);
		int ans = 0;
		for (int j = 1; j <= n; ++j)
			if (!vis[j])
				++ans;
		printf("%d\n", ans);
	}
	return 0;
}

Answer is here!

#include<bits/stdc++.h>
using namespace std;
int s[114][514],d[114];
bool o[114];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>d[i];
        for(int j=1;j<=d[i];j++)cin>>s[i][j];
    }
    int m;
    cin>>m;
    while(m--)
    {
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            cin>>o[i];
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=d[i];j++)
            {
                if(!o[s[i][j]])o[i]=0;
            }
        }
        for(int i=1;i<=n;i++)if(o[i])ans++;
        cout<<ans<<endl;
    }
}

Thank you very much!