连通块中点的数量

这个题很简单，只要在原来并查集模板的基础上增加一个记录点的数量的数组，根本不用把它当图做。

#include <bits/stdc++.h>
using namespace std;
int p[100005],s[100005],n,m;
int find(int x){
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int main(){
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) p[i] = i,s[i] = 1;
    while(m --){
        string t;
        int a,b;
        cin >> t;
        if(t == "C"){
            cin >> a >> b;
            a = find(a),b = find(b);
            if(a == b) continue;
            p[a] = b;
            s[b] += s[a];//将两个集合点的数量加起来
        }else if(t == "Q1"){
            cin >> a >> b;
            a = find(a),b = find(b);
            if(a == b){
                cout << "Yes" << endl;
            }else{
                cout << "No" << endl;
            }
        }else{
            cin >> a;
            a = find(a);
            cout << s[a] << endl;
        }
    }
    return 0;
}