（1）朴素并查集：P39

存储每个点的祖宗节点
int p[N];

返回x的祖宗节点

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

初始化，假定节点编号1~n
for (int i = 1; i <= n; i ++) p[i] = i;

合并a和b所在的两个集合
p[find(a)] = find(b);

#include <iostream>

using namespace std;

const int N = 100010;
int n, m;
int p[N];

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

int main () {
    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[0] == 'M') p[find(a)] = find(b);
        else {
            if (find(a) == find(b)) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}