@ jalaj consider a graph 2->1 3->1 2->3
apply dfs(1) n dfs(2)... will give u 2 components although GRAPH is connected.... On Fri, Jul 9, 2010 at 10:23 PM, jalaj jaiswal <[email protected]>wrote: > > int count=0; > for every vertex v{ > if visited[v]==0 > dfs(v) > count++; > } > count is the number of components in graph........ or m i missing something > ? > > > On Fri, Jul 9, 2010 at 9:23 PM, amit <[email protected]> wrote: > >> How to check if a directed graph is connected. >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]<algogeeks%[email protected]> >> . >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> >> > > > -- > With Regards, > Jalaj Jaiswal > +919026283397 > B.TECH IT > IIIT ALLAHABAD > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<algogeeks%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
