At the each level, traversed by BFS, you will have to check whether the vertex in this level has the element same as it found in the previous level. If it is different, then count it.
On Sun, May 29, 2011 at 10:43 PM, anshu mishra <[email protected]>wrote: > @piyush > > void bfs(int mat[][n], bool flag[][n], int i, int j) > { > queue.push(mat[i][j]); > while (!q.empty()) > { > x = q.top(); > q.pop(); > add top bottom, left right element in qeuue if their flag is true and their > value is equal to x and mark their flag false; > } > } > > -- > 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. > -- -Aakash Johari (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]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
