it can also be done if we calculate values diagonal wise in the matrix.

At each step choose maximum value , at last we'll get maximum value.



Sanju
:)



On Sun, Aug 21, 2011 at 6:59 PM, sarvesh saran
<[email protected]>wrote:

> 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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>>
>>
>> --
>>
>>
>> Nagarajan S
>>
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