@anand: the greedy appoach will not work in this test case:
{2,10,25,
1,20,10,
100, 5, 100}
i think use DP. create an 'max matrix of same order.
max[i,j]= maximum( max[i-1][j] + a[i][j], max[i][j-1]+a[i][j] );
max[n-1][n-1] will be the answer.
Tell me if i m wrong.
On Oct 2, 10:29 pm, Anand <[email protected]> wrote:
> Since in our case starting point is fixed "ie top left corner" so dynamic
> programmng will not make any difference. Dynamic programing makes difference
> only when starting point is not fixed. Solution from Greedy and Dynamic
> programming will be same in this case. Correct me if I am wrong
>
> On Sat, Oct 2, 2010 at 8:27 AM, Mridul Malpani <[email protected]>wrote:
>
> > @ anand: the code u have given is an greedy approach. & it will not
> > work.
>
> > On Oct 1, 12:34 am, Anand <[email protected]> wrote:
> > > Here is a code for solving the problem using DP.
> >http://codepad.org/AoPtCmwA
>
> > > On Thu, Sep 30, 2010 at 3:01 AM, Modeling Expert <
>
> > > [email protected]> wrote:
> > > > recurssion...
>
> > > > At any point X
>
> > > > val_t getMax( position X){
>
> > > > ( ! End of Table )
> > > > sum = GetApples[X] + MAX ( getMax(X_down) , getMax
> > > > ( X_right) ) ;
>
> > > > returnn sum ;
> > > > }
>
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