hdoj--5569--matrix(动态规划)

matrix

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 509    Accepted Submission(s): 297

Problem Description

Given a matrix with 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

Several test cases(about 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
  
  

 

Source

BestCoder Round #63 (div.2)

 

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用dp数组记录下到达每一个点最小的路径

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long long dp[1010][1010],num[1010][1010];
int m,n;
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(num,0,sizeof(num));
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		scanf("%lld",&num[i][j]);
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				if(i==1&&j==1)
				dp[i][j]=0;
				else if((i+j)&1)
				{
					if(i==1)
					dp[i][j]=dp[i][j-1]+num[i][j-1]*num[i][j];
					else if(j==1)
					dp[i][j]=dp[i-1][j]+num[i-1][j]*num[i][j];
					else
					dp[i][j]=min(dp[i][j-1]+num[i][j-1]*num[i][j],dp[i-1][j]+num[i-1][j]*num[i][j]);
				}
				else
				{
					if(i==1)
					dp[i][j]=dp[i][j-1];
					else if(j==1)
					dp[i][j]=dp[i-1][j];
					else dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
				}
			}
		}
		printf("%lld\n",dp[n][m]);
	}
	return 0;
}