陈老师的多校联合20140816A题||spoj 10228 动态规划

http://www.spoj.com/problems/AMR11A/

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

 

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

题目大意：给定一个n*m的棋盘，（1,1）和（n，m）位置为0，现在一人从左上往右下角走每次只能走(i,j+1),(i+1,j)，并且下一步的值是上一步值+a[i][j],并且下一步的值必须为正数，走到最后。应给初始位置一个什么样的值才能使条件成立的情况下走到终点。

解题思路：我们用dp[i][j]表该点活下来需要的最小值，现在知道dp[n][m]的值必为1，状态转移方程：

              dp[i-1][j]=min(dp[i-1][j],dp[i][j]-a[i-1][j]);
               if(dp[i-1][j]<=0)
                  dp[i-1][j]=1;
               dp[i][j-1]=min(dp[i][j-1],dp[i][j]-a[i][j-1]);
               if(dp[i][j-1]<=0)
                  dp[i][j-1]=1;.

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int a[550][555],dp[555][555];
int n,m;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        memset(dp,0x3f3f3f,sizeof(dp));
        dp[n][m]=1;
        for(int i=n;i>0;i--)
           for(int j=m;j>0;j--)
           {
               dp[i-1][j]=min(dp[i-1][j],dp[i][j]-a[i-1][j]);
               if(dp[i-1][j]<=0)
                  dp[i-1][j]=1;
               dp[i][j-1]=min(dp[i][j-1],dp[i][j]-a[i][j-1]);
               if(dp[i][j-1]<=0)
                  dp[i][j-1]=1;
           }
        printf("%d\n",dp[1][1]);
    }
    return 0;
}