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  • 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 a1a2+a3a4+...+a2k1a2k. What is the minimum of the cost?
     

    Input
    Several test cases(about 5)

    For each cases, first come 2 integers, n,m(1n1000,1m1000)

    N+m is an odd number.

    Then follows n lines with m numbers ai,j(1ai100)
     

    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
     

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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
    ",dp[n][m]);
    	}
    	return 0;
    }

     
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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273670.html
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