Hi Nagaraj,

Should it not be:

(a[i+1][j],a[i][j+1])+a[i][j] ? That is given (i,j) find max of next 2
possible cells?

Why are we subtracting instead of adding?


thanks,
Sarvesh

On Mon, Aug 22, 2011 at 7:11 AM, nagarajan <[email protected]> wrote:

> Just its a matrix manipulation .. similar to knap sack problem...
>
>
> i am filling each value of matrix by finding the max
> (a[i-1][j],a[i][j-1])+a[i][j] ..
>
>
> so that the max value will be stored in the each index  positions and so in
> the a[M][N]..
>
>
> On Mon, Aug 22, 2011 at 1:34 AM, sarvesh saran <[email protected]
> > wrote:
>
>> 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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>>
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>
>
>
> --
>
>
> Nagarajan S
>
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