非常好的一道练习并查集的题目，逆向思维很巧妙

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int M = 5e4 + 10;
const int D = 1e3+10;
typedef pair<int, int>PII;
vector<int>a[M];//用来存边
int n, m, d;
int lock[D];//用来存第i天封锁的机场编号
int p[M];
int ans[D];//存第i天的答案
int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> d;
    vector<PII>query[D];//用来存第i天的所有查询
    for(int i=1;i<=n;i++) p[i] = i;
    for (int i = 0; i < m; i++) {
        int s, t;
        cin >> s >> t;
        a[s].push_back(t);
        a[t].push_back(s);
    }
    for (int i = 1; i <= d; i++) {
        int s,Q;
        cin >>s>> Q;
        lock[i] = s;
        p[s] = 0;
        vector<PII> temp;
        for (int j = 0; j < Q; j++) {
            int w, t;
            cin >> w >> t;
            temp.push_back({w,t});
        }
        query[i] = temp;
    }

    //逆向思维，从最后一天开始，依次将被封锁的航班的所有邻边进行一次union,但是会出现将已经封锁的边也union的情况，
    //需要进行判断
    for (int i = 1; i <= n; i++) {
        if (p[i]) {
            for (int j = 0;j<a[i].size();j++){
                int t = a[i][j];
                if(p[t]) p[find(t)] = find(i);
            }
        }
    }
    for (int i = d; i >= 1; i--) {
        for (int j = 0; j < query[i].size(); j++) {
            auto t = query[i][j];
            if (find(t.first)!= find(t.second)|| find(t.first) == 0 || find(t.second) == 0) ans[i]++;
        }
        int s = lock[i];
        p[s] =s;
        for (int j = 0; j < a[s].size(); j++) {
            int t = a[s][j];//将lock[i]的所有非封锁邻边union
            if (p[t]) p[find(s)]= find(t);
        }
    }
    for (int i = 1; i <= d; i++) cout << ans[i]<<endl;
    return 0;
}