A simple Cycle detection Problem using DFS :: Codeforces 977E

 Problem Link

In this problem several connected components are given, we need to find number of such components with cycle.

A components is contain a cycle if and only if all of its vertices has degree exactly 2.

To find connected components of the graph, we will use  a simple dfs through unvisited nodes. You can learn dfs from here

After that we will check if all the vertices has degree exactly 2 or not.

Official Editorial

Code:

 

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#include<bits/stdc++.h> using namespace std; #define pb push_back ///the connected component is a cycle iff the degree of each its vertex equals to 2 const int maxn=200005; vector<int>edge[maxn],connected; int vis[maxn],degree[maxn]; int cnt=0; void dfs(int u){ vis[u]=1; connected.pb(u); for(auto it:edge[u]){ if(vis[it]==0){ dfs(it); } } } int main(){ int n,m; cin>>n>>m; for(int i=0;i<m;i++){ int u,v; cin>>u>>v; edge[u].pb(v); edge[v].pb(u); degree[u]++; degree[v]++; } for(int i=1;i<=n;i++){ if(vis[i]==0){ connected.clear(); dfs(i); bool ok=1; for(auto it:connected){ if(degree[it]!=2){ ok=0; } } if(ok){ cnt++; } } } cout << cnt << endl; }