牛客训练联盟热身训练赛第二场补题报告

F.Interstellar Love

1.思路

   并查集求无向图连通分量和闭环数：连通分量用判断祖先节点是否等于自身节点求出，两个节点的祖先节点相同且又添加了新边，则闭环数++。

2.代码

 

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int fa[10005];
int vis[10005];
int sig[10005];
int find(int x) {
    if(fa[x] == x) {
        return x;
    } else {
        return fa[x] = find(fa[x]);
    }
}
void merge(int x, int y) {
    int fx = find(x);
    int fy = find(y);
    if(fx != fy) {
        fa[fx] = fy;
        if(sig[fx]) {
            sig[fy] = 1;
        }
    } else {
        sig[fx] = 1;
    }
}
int main() {
    int t;
    cin >> t;
    for(int i = 1; i <= t; i++) {
        int n, m;
        cin >> n >> m;
        for(int j = 0; j <= n; j++) {
            fa[j] = j;
            vis[j] = 0;
            sig[j] = 0;
        }
        int x, y;
        for(int j = 0; j < m; j++) {
            cin >> x >> y;
            vis[x] = 1;
            vis[y] = 1;
            merge(x, y); 
        }
        int cnt = 0;
        int huan = 0;
        for(int j = 1; j <= n; j++) {
            if(j == fa[j] && vis[j]) {
                cnt++;
                if(sig[j] == 1) {
                    huan++;
                }
            }
        }
        printf("Night sky #%d: %d constellations, of which %d need to be fixed.\n\n", i, cnt, huan);
    }

    return 0;
}