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题目

837. 连通块中点的数量

思路

与上一题类似，多了一个统计集合数量，可以定义一个数组用于记录集合个数，初始时设置为1，每要合并就更新一次。

代码

#include<iostream>
using namespace std;
const int N = 100010;
int n, m;
int p[N],siz[N];
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;siz[i] = 1;}while (m--){char op[3];cin >> op;int a, b;if (op[0] == 'C'){cin >> a >> b;if (find(a) == find(b))continue;siz[find(b)] = siz[find(a)] + siz[find(b)];p[find(a)] = find(b);}else if (op[1] == '1'){cin >> a >> b;if (find(a) == find(b))cout << "Yes" << endl;elsecout << "No" << endl;}else {cin >>a;cout << siz[find(a)] << endl;}}return 0;
}