原题目：森林里的鸟
https://www.acwing.com/solution/content/184005/
并查集模板包含两部分，find和merge，find包含了路径压缩，merge包含了启发式合并。
启发式合并需要一个sz数组用于判断以它为根节点的子树深度。

#include <iostream>
#include <unordered_set>
using namespace std;

const int N = 1e5 + 10;
int p[N], s[N];
int n;
unordered_set<int> birds, tr;

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

void merge(int x, int y) {
    int a = find(x), b = find(y);
    if (a == b)return;
    if (s[a] < s[b])swap(a, b);
    p[b] = a;
    s[a] += s[b];
}

int main() {
    for (int i = 0; i < N; i++) {
        p[i] = i;
        s[i] = 1;
    }
    cin >> n;
    while (n--) {
        int k, b, t;
        cin >> k >> b;
        birds.insert(b);
        for (int i = 1; i < k; i++) {
            cin >> t;
            merge(b, t);
            birds.insert(t);
        }
    }
    int q;
    cin >> q;
    for (auto it = birds.begin(); it != birds.end(); it++) {
        if (tr.find(*it) == tr.end()) {
            tr.insert(find(*it));
        }
    }
    cout << tr.size() << ' ' << birds.size() << '\n';
    while (q--) {
        int x, y;
        cin >> x >> y;
        if (find(x) == find(y))
            cout << "Yes\n";
        else cout << "No\n";
    }
}