hdoj matrix 5569 （奇偶DP） 好题 向右下走取使方程值最小

matrix

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 107    Accepted Submission(s): 73

Problem Description

n rows and m columns ( n+m is an odd number ), at first , you begin with the number at top-left corner (1,1) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost is a1∗a2+a3∗a4+...+a2k−1∗a2k. What is the minimum of the cost?

Input

5)

For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000)

N+m is an odd number.

Then follows n lines with m numbers ai,j(1≤ai≤100)

Output

For each cases, please output an integer in a line as the answer.

Sample Input

2 3
1 2 3
2 2 1
2 3
2 2 1
1 2 4

Sample Output

4 8

#include<stdio.h>
#include<string.h>
int min(int x,int y)
{
	return x>y?y:x;
}
int a[1010][1010];
int dp[1010][1010];
int main()
{
	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
			for(j=1;j<=m;j++)
				scanf("%d",&a[i][j]);
		memset(dp,0x3f3f3f,sizeof(dp));
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				if((i+j)&1)//奇数步时考虑 
					dp[i][j]=min(dp[i-1][j]+a[i-1][j]*a[i][j],dp[i][j-1]+a[i][j-1]*a[i][j]);
				else
					dp[i][j]=min(dp[i-1][j],dp[i][j-1]);//偶数步时，取向右走，与向下走的最小值。 
				if(i==1&&j==1)
					dp[i][j]=0;
			}
		}
		printf("%d\n",dp[n][m]);
	}
	return 0;
}