@ross : * if their flag is true and their value is equal to x*

On Sun, May 29, 2011 at 11:07 PM, ross <[email protected]> wrote:

> @Anshu
> If you do
> " add top bottom, left right element to the popped element in qeuue "
> should you need to do it for each element in the matrix.
> So, will that not be O(n3)??
>
> Clarify if i am wrong.
>
> On May 30, 9:52 am, Aakash Johari <[email protected]> wrote:
> > 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;
> > > }
> > > }
> >
> > > --
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> >
> > --
> > -Aakash Johari
> > (IIIT Allahabad)
>
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>


-- 
-Aakash Johari
(IIIT Allahabad)

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