as long as it is an undirected graph;simple dfs would suffice! On Tue, Nov 24, 2009 at 10:13 PM, Rohit Saraf <[email protected]>wrote:
> But then .. > think.. > for every two pairs of nodes you have to check whether both are connected > to root and then you will say yes/no. > And you will have to do this for all disjoint components. > It's right but not O(n^2). > > -- > 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. > -- Thanks, Chakravarthi. -- 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.
