836. 合并集合 题解

#include <iostream>
using namespace std;
const int N = 100010;

int n, m, a, b;
int p[N];
string op;

int find(int x)  // 带路径压缩的查询
{
    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;

    while (m -- )
    {
        cin >> op >> a >> b;
        if (op == "M") p[find(a)] = find(b);  // 合并集合
        else 
        {
            if (find(a) == find(b)) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}