You just need to run DFS once. It will give you all the connected components. So it is indeed O(n+e)
On Tue, Nov 24, 2009 at 2:43 PM, Rohit Saraf <rohit.kumar.sa...@gmail.com>wrote: > Isnt it running DFS on every node? > 1) Pick any node in the graph say n > 2) With n as root, run DFS. While running DFS, if you visit node x, mark > M[n,x]=1 and M[x,n]=1 . > The running time is O(k+E) [not k, you are forgetting the case when E != > O(n)] .Also have a hash finished[n]=true. > 3) Repeat steps 1-2 for each node until finished[node] = true for all of > them. > > So overall time is not O(n^2). A proof of O(n^2) is also reqd. with the > solution > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > 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 algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
