CareerCup

warshall transitive closure

hey guys ..i think we can do by des steps.
1) build depth first trees
2) at each step find the next neighbour for those node which are connected to previous one and d the same till n.

and for undirected graphs...?

Do we really need transitive closures for this one?. I mean i=the question is asking about nth neighbor..so cant we extract that information from adjacency matrix itself?
A B C D . . Z
A
B
C
D
.
.
.Z

So, for each row, we need to find out the nth entry that has a '1' marked on it and if it is available then return the corresponding column vertex.

Also, I dont see any modification needed for directed graph.

In case im mistaken,...please do correct me!.

Thanks,
Pavan.