# SPOJ - AMR11A（DP）

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 top-left corner cell (1,1) and the Sorcerer's Stone is in the bottom-right 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

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

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.

 1 #include<iostream>
2 #include<cstdlib>
3 #include<cstring>
4 #include<cstdio>
5 #include<string>
6 #include<cmath>
7 #include<algorithm>
8 #include<stack>
9 #include<queue>
10 #define ll long long
11 #define inf 0x3f3f3f3f
12 #define pi 3.141592653589793238462643383279
13 using namespace std;
14 int t,n,m,map[505][505],dp[505][505];
15 int main()
16 {
17     cin>>t;
18     while(t--)
19     {
20         cin>>n>>m;
21         for(int i=1; i<=n; ++i)
22             for(int j=1; j<=m;++j)
23                 scanf("%d",&map[i][j]);
24
25         dp[n][m] = 1;
26         for(int i=n-1; i>=1; --i)
27             dp[i][m] = max(1,dp[i+1][m] - map[i][m]); //逆推最后一列的dp值
28
29         for(int i=m-1; i>=1; --i)
30             dp[n][i] = max(1,dp[n][i+1] - map[n][i]); //逆推最后一行的dp值
31
32         for(int i=n-1; i>=1; --i)
33             for(int j=m-1; j>=1; --j)
34                 dp[i][j] = max(1,min(dp[i+1][j], dp[i][j+1]) - map[i][j]); //逆推剩余点的dp值
35         //只能先推最后一行和最后一列，因为dp值的推导必须是连续的，你当前的dp值必须由已经知晓的dp值推出
36         cout<<dp[1][1]<<endl;
37     }
38     return 0;
39 }

posted @ 2018-07-10 09:45  特务依昂  阅读(173)  评论(0编辑  收藏  举报