CodeForces - 977E Cyclic Components dfs

题意:

找到所有的单链环的个数

思路:

dfs 遍历整个图,每次把一个连通块标记出来,并判读是否每个点的度数都为2(是否为单链环);


#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 2e5+7;
int n, m, f;
bool vis[maxn];
vector<int> vec[maxn];
void dfs(int id) {
    vis[id]=1;
    if(vec[id].size() != 2) f = 0;
    for(auto i : vec[id]){
        if(vis[i]) continue;
        dfs(i);
    }
    return;
}
int main() {
    cin >> n >> m;
    for(int i = 0; i < m; ++i) {
        int u, v;
        cin>>u>>v;
        vec[u].push_back(v);
        vec[v].push_back(u);
    }
    int ans = 0;
    for(int i = 1; i <= n; ++i){
        if(!vis[i]) {
            f = 1;
            dfs(i);
            if(f) ans++;
        }
    }
    cout<<ans;
    return 0;
}














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转载自blog.csdn.net/xiang_6/article/details/80385409