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
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souvikd_pro:
20190919 19:29:23
memoization also worked !! 

quannguyenlhp:
20190813 03:49:14
AC in one billionth go. 

ballerbuoy:
20190801 05:18:31
Woah! AC in one go!


g_unit:
20190704 18:10:51
guys dont forget to reinitialize your dp array and reinitialize the variable you are using to store the maximum value.. ac in second go! 

yogesh1208:
20190618 22:55:04
it is giving nzec in java


saurav_paul:
20190527 16:25:13
Accepted in one go.


tarusus:
20190520 15:48:10
Don't think too much> simple iterative approach work.


scolar_fuad:
20190510 18:40:39
I used bottom up dp


medha_08:
20190429 18:42:38
AC in one go :) 

syed_tanveer:
20190425 09:54:17
Easy one, O(h*w) passed in 0.02 sec. 
Added by:  Paritosh Aggarwal 
Date:  20090221 
Time limit:  0.141s0.393s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
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