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
leafbebop:
20170601 20:11:40
It would be a good problem, if only golang is permitted. Last edit: 20170601 20:12:20 

Nafis Islam:
20170527 09:14:54
AC my 26th :D 

da_201501181:
20170526 07:57:01
AC in one go..!! java 0.1 s 

sagnik_66:
20170524 16:08:08
3rd DP problem and 1st AC in one go!! Helps build confidence. Bottomup!! Last edit: 20170524 16:08:33 

horizon121:
20170516 16:42:09
AC in one go..Honestly its very simple 

vinhvo:
20170309 22:19:09
Not worth solving, malformed input ! 

anmol23:
20170305 21:46:20
I solved by bottom up. Can anyone give a clue how to solve by top down? 

abinator_1308:
20170302 18:38:40
Bottom's up !!!


vladimira:
20170221 12:31:12
Dozens NZEC with python.


nilabja16180:
20170215 15:51:20
AC in one GO! nd Dp!! 
Added by:  Paritosh Aggarwal 
Date:  20090221 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  ADA95 ASM32 BASH BF C CSHARP CPP C99 CLPS LISP sbcl LISP clisp D FORTRAN HASK ICON ICK JAVA LUA NEM NICE OCAML PASGPC PASFPC PERL PHP PIKE PRLGswi PYTHON RUBY SCM qobi SCM guile ST TEXT WHITESPACE 