BYTESM2  Philosophers Stone
One of the secret chambers in Hogwarts is full of philosopher’s stones. The floor of the chamber is covered by h × w square tiles, where there are h rows of tiles from front (first row) to back (last row) and w columns of tiles from left to right. Each tile has 1 to 100 stones on it. Harry has to grab as many philosopher’s stones as possible, subject to the following restrictions:
 He starts by choosing any tile in the first row, and collects the philosopher’s stones on that tile. Then, he moves to a tile in the next row, collects the philosopher’s stones on the tile, and so on until he reaches the last row.
 When he moves from one tile to a tile in the next row, he can only move to the tile just below it or diagonally to the left or right.
Input
The first line consists of a single integer T, the number of test cases. In each of the test cases, the first line has two integers. The first integer h (1 <= h <= 100) is the number of rows of tiles on the floor. The second integer w (1 <= w <= 100) is the number of columns of tiles on the floor. Next, there are h lines of inputs. The ith line of these, specifies the number of philosopher’s stones in each tile of the ith row from the front. Each line has w integers, where each integer m (0 <= m <= 100) is the number of philosopher’s stones on that tile. The integers are separated by a space character.
Output
The output should consist of T lines, (1 <= T <= 100), one for each test case. Each line consists of a single integer, which is the maximum possible number of philosopher’s stones Harry can grab, in one single trip from the first row to the last row for the corresponding test case.
Example
Input: 1 6 5 3 1 7 4 2 2 1 3 1 1 1 2 2 1 8 2 2 1 5 3 2 1 4 4 4 5 2 7 5 1 Output: 32 //7+1+8+5+4+7=32
hide comments
amarveer:
20141027 18:32:03
My first dp :)


jetpack:
20141016 21:21:10
first dp.^_^ 

manu4rhyme:
20140922 19:10:12
easy dp..Accepted in one go :) 

Deepanker Aggarwal:
20140920 14:13:18
Someone, please enable c++4.3.2 . I don't think it changes the complexity of the question 

Sumit Gulati:
20140919 11:29:21
nice question of dp for practise :) 

Rajat (1307086):
20140917 23:06:50
need to practice more DP. 

shade_1:
20140917 20:58:15
easy problm...should be moved to tutorial section


Ruffneck:
20140916 23:16:13
this question cleared my concepts on DP. easy, but awesome. 

Gaurav Ahirwar:
20140905 13:31:10
AC! in one go! :D .. Finally improving my programming! 

Pratik Nagelia:
20140806 22:19:15
AC in one go... :)B)..!

Added by:  Paritosh Aggarwal 
Date:  20090221 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
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