并查集

//indices指标,指数  continuously连续不断地  numbered编号的给…编号  no more than至多，不超过
//which is the number of queries. Then Q lines follow, each contains the indices of two birds.
//这是查询的数量.接下来是Q行，每行都包含两只鸟的索引.

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

const int N = 10010;

int n;
int p[N];
int birds[10];
bool st[N];

int find(int x)
{
    if(p[x]!=x) p[x] = find(p[x]);
    return p[x];
}

int main()
{
    cin>>n;
    for(int i = 1; i<N; i++) p[i] = i;

    int cnt = 0;//合并多少次
    for(int i = 0; i<n; i++)
    {
        int k;
        cin>>k;
        for(int j = 0; j<k; j++)
        {
            cin>>birds[j];
            st[birds[j]] = true;
        }

        for(int j = 1; j<k; j++)
        {
            int a = birds[j-1],b = birds[j];
            a = find(a),b = find(b);
            if(a!=b)
            {
                p[a] = b;
                cnt++;
            }
        }
    }
    int sum = 0;
    for(int i = 0; i<N; i++) sum += st[i];//鸟的总数
    printf("%d %d\n",sum-cnt,sum);//树的数量=鸟的总数-合并的次数

    int q;
    cin>>q;
    while(q--)
    {
        int a,b;
        cin>>a>>b;
        if(find(a)==find(b)) puts("Yes");
        else puts("No");
    }
    return 0;
}