陳老師的多校聯合20140816A題||spoj 10228 動態規劃

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

題目大意：給定一個n*m的棋盤，（1,1）和（n，m）位置為0，現在一人從左上往右下角走每次只能走(i,j+1),(i+1,j)，並且下一步的值是上一步值+a[i][j],並且下一步的值必須為正數，走到最後。應給初始位置一個什麼樣的值才能使條件成立的情況下走到終點。

解題思路：我們用dp[i][j]表該點活下來需要的最小值，現在知道dp[n][m]的值必為1，狀態轉移方程：

              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;
}