I didnt understand the above solution, but I have a bad complexity solution. Make an nXn boolean adjacency matrix. Run transitivy closure on the matrix. For every node w, check if for all v E V, w->v=> v->w
On Oct 13, 9:52 pm, praba karan <prabaf...@gmail.com> wrote: > spoj.pl/problems/BOTTOM > > my algo for this prob goes by this way..will this work correctly? > > 1)first make a bfs from an arbitrary node and store the nodes visited > in order in an array. > > 2)the node that is visited at the end of the BFS in the 1st bfs is now > taken as the root and then make a BFS from this node and store the > nodes visited in order in an array. > > 3)reverse the array in step 2 > > 4)find the Longest Common Subsequence between the arrays in step 1 and > step 3 > > 5)if any one of the array is empty,then all the other nodes in the > other array sud b printed. > > will this logic work? > > any other ideas are welcomed... > > praba... -- 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.
