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
ducky94tb:
20171222 04:14:30
BufferedReader works fine !! NZEC because of bad input format.


floofybooper:
20171217 05:49:36
AC using middle center! 

oldpig2017:
20171113 05:34:51
The input is malformed. Don't waste your time on this problem. 

sinersnvrsleep:
20171111 14:27:01
AC in one go no dp required !!! 

g_o_d:
20171020 18:54:30
DP :) !! 

aman22222:
20171020 09:57:27
Good Confidence builder! 

fadhilmusaad:
20171012 10:27:41
Last edit: 20171012 10:28:39 

vishesh197:
20170918 12:27:19
just use dp approach if tle is there.State of dp is maximum path to reach a tile(dp[i][j]).AC in third go.


nwarrior:
20170917 12:16:31
Accepted in one go


rohit9934:
20170810 09:59:16
Last edit: 20180311 09:49:00 
Added by:  Paritosh Aggarwal 
Date:  20090221 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
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