AMR11A  Magic Grid
Thanks a lot for helping Harry Potter in finding the Sorcerer's Stone of Immortality in October. Did we not tell you that it was just an online game ? uhhh! now here is the real onsite task for Harry. You are given a magrid S ( a magic grid ) having R rows and C columns. Each cell in this magrid has either a Hungarian horntail dragon that our intrepid hero has to defeat, or a flask of magic potion that his teacher Snape has left for him. A dragon at a cell (i,j) takes away S[i][j] strength points from him, and a potion at a cell (i,j) increases Harry's strength by S[i][j]. If his strength drops to 0 or less at any point during his journey, Harry dies, and no magical stone can revive him.
Harry starts from the topleft corner cell (1,1) and the Sorcerer's Stone is in the bottomright corner cell (R,C). From a cell (i,j), Harry can only move either one cell down or right i.e., to cell (i+1,j) or cell (i,j+1) and he can not move outside the magrid. Harry has used magic before starting his journey to determine which cell contains what, but lacks the basic simple mathematical skill to determine what minimum strength he needs to start with to collect the Sorcerer's Stone. Please help him once again.
Input (STDIN):
The first line contains the number of test cases T. T cases follow. Each test case consists of R C in the first line followed by the description of the grid in R lines, each containing C integers. Rows are numbered 1 to R from top to bottom and columns are numbered 1 to C from left to right. Cells with S[i][j] < 0 contain dragons, others contain magic potions.
Output (STDOUT):
Output T lines, one for each case containing the minimum strength Harry should start with from the cell (1,1) to have a positive strength through out his journey to the cell (R,C).
Constraints:
1 ≤ T ≤ 5
2 ≤ R, C ≤ 500
10^3 ≤ S[i][j] ≤ 10^3
S[1][1] = S[R][C] = 0
Sample Input:
3
2 3
0 1 3
1 2 0
2 2
0 1
2 0
3 4
0 2 3 1
1 4 0 2
1 2 3 0
Sample Output:
2
1
2
Explanation:
Case 1 : If Harry starts with strength = 1 at cell (1,1), he cannot maintain a positive strength in any possible path. He needs at least strength = 2 initially.
Case 2 : Note that to start from (1,1) he needs at least strength = 1.
hide comments
vidz_1:
20200924 14:46:48
Why can't we just do normal forward dp and make the table. Later we go from last block to initial block always going to the best path(max(dp[i1][j],dp[i][j1]) because that's how the table would be built, find the min value and return mod(min)+1 as the answer ? Can someone please help, I am getting a WA. 

tokyoghoul1993:
20191229 10:12:16
Can someone plz explain how binary search is used here?


yashranjan74:
20190917 22:44:03
Nice problem. Can be easily done with any way, TopDown DP or BottomUp DP. 

chirayu_555:
20190814 21:03:12
AC in one go..!! Nice problem..!!


geekymanas_:
20190729 14:17:17
There are two ways to do this. First is moving forward dp with Binary Search, second moving backward just dp Last edit: 20190729 18:57:29 

Asokan R:
20190104 21:47:39
Bottom up DP in Java gives TLE, but same code in c++ is accepted Last edit: 20190104 21:47:55 

hello_world123:
20180714 21:50:06
Keep in mind that initial strength is minimum 1. 

shahi9935:
20180626 05:28:32
Thanks @sherlock11 Last edit: 20180626 05:29:02 

sherlock11:
20180623 15:02:31
its not much of dp problem rather a variation of binary search................... 

jack_greene999:
20180623 14:09:49
AC in one go 
Added by:  Varun Jalan 
Date:  20111215 
Time limit:  1s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 
Resource:  Anil Kishore  ICPC Asia regionals, Amritapuri 2011 