Cyclic Components CodeForces - 977E

Cyclic Components CodeForces - 977E

分析：DFS 如果其中一个连通图的所有点的度数都为2就符合题意(搜索完某一连通图就把该连通图的所有点做标记，不再访问)

代码：

#include<bits/stdc++.h>
using namespace std;
const int N = 200000 + 5; 
vector<int> a[N];
int vis[N];
int flag = 1;
void dfs(int cur) {
    vis[cur] = 1;
    if(a[cur].size() != 2) flag = 0;
    for(int i : a[cur]) {
        if(!vis[i]) dfs(i);
    }
} 
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    int x, y;
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &x, &y);
        a[x].push_back(y);
        a[y].push_back(x);
    }
    memset(vis, 0, sizeof(vis));
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        flag = 1;    
        if(!vis[i]) {
            dfs(i);
            if(flag) ans++;
        }
    }
    printf("%d\n", ans);
    return 0;
}