A simple DFS will do good.A 2D vector is used to represent the graph via adjacency list.
All you have to do is have a visited array to mark nodes in the graph that is already visited. Have a count variable to count the number of groups available.
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| #include<iostream> | |
| #include<cstdio> | |
| #include<vector> | |
| #include<cstring> | |
| using namespace std; | |
| bool visited[100005]; | |
| vector<vector<int> > adj(100005); | |
| int n; | |
| void dfs(int i){ | |
| if(visited[i]) | |
| return; | |
| visited[i]=1; | |
| for(vector<int>::iterator it=adj[i].begin();it!=adj[i].end();it++){ | |
| if(!visited[*it]) | |
| dfs(*it); | |
| } | |
| } | |
| int main(){ | |
| int t; | |
| scanf("%d",&t); | |
| while(t--){ | |
| int e,ans=0,x,y; | |
| scanf("%d %d",&n,&e); | |
| memset(visited,0,sizeof visited); | |
| for(int i=0;i<n;i++) | |
| adj[i].clear(); | |
| for(int i=0;i<e;i++){ | |
| scanf("%d %d",&x,&y); | |
| adj[x].push_back(y); | |
| adj[y].push_back(x); | |
| } | |
| for(int i=0;i<n;i++){ | |
| if(!visited[i]){ | |
| ans++; | |
| dfs(i); | |
| } | |
| } | |
| printf("\n%d",ans); | |
| } | |
| } | |
coding hangover 