Hi Nagaraj,

Can you explain the logic behind your code? it will be a great help!

thanks,
Sarvesh



On Mon, Aug 22, 2011 at 1:21 AM, nagarajan <[email protected]> wrote:

> it can be done in the following way,
>
>
> let the input is in the array a[N+1][M+1]
>
> we process the input as
>
> 0 0 0 0 0
> 0 2 2 3 0
> 0 0 3 1 1
> 0 1 2 2 1
> 0 4 1 2 2
>
> compute the matrix as
>
> a[i][j] = max(a[i-1][j], a[i][j-1])+a[i][j]; for i=1 to N and j=1 to M
>
> you will get the required result in a[N][M]
>
> the answer looks as
>
> 0 0 0 0 0
> 0 2 4 7 7
> 0 2 7 8 9
> 0 3 9 11 12
> 0 7 10 13 15
>
> i think this is right... any mistakes in this , you are welcome..
>
> On Mon, Aug 22, 2011 at 1:00 AM, sarvesh saran <[email protected]
> > wrote:
>
>> HI,
>>
>> I have a classic chessboard and rice problem. the detailed statement is in
>> the link:http://pastebin.com/nykkFP84
>>
>> my solution is :
>>
>>    1. int max_rice(std::vector<std::vector<int>> const& board, unsigned
>>    x, unsigned y)
>>    2. {
>>    3. if (y == board.size() || x == board[0].size()) return 0;
>>    4.
>>    5. return std::max(max_rice(board, x + 1, y), max_rice(board, x, y +
>>    1)) + board[y][x];
>>    6. }
>>
>> I would like to find a dynamic programming solution to this problem. Any
>> ideas?
>>
>> thanks,
>> Sarvesh
>>
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>
>
>
> --
>
>
> Nagarajan S
>
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