AcWing 836. 合并集合

#include <iostream>
using namespace std;
const int N = 100010;
int p[N];
int find(int x) {//返回祖宗节点，同时进行路径压缩
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) p[i] = i;
    while (m -- ) {
        char op[2];
        int a, b;
        scanf("%s%d%d", op, &a, &b);
        if (*op == 'M') p[find(a)] = find(b);
        else {
            if (find(a) == find(b)) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}