matrix

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5569

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

HINT

题意

给定n*m(n+m为奇数)的矩阵，从（1,1）走到(n,m)且只能往右往下走，设经过的数为a1,a2,..,a2k，贡献为a1*a2+a3*a4...+a2k-1*a2k，求最小贡献

题解：

dp[i][j]表示走到i,j的最小贡献，我们只考虑(i+j)为奇数的时候就好了

然后转移的时候，也只会从奇数位置转移过来

代码：

#include<iostream>
#include<cstring>
#include<stdio.h>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
int n,m;
int mp[1005][1005];
int dp[1005][1005];
const int inf = 1e9+5;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<=n;i++)
            for(int j=0;j<=m;j++)
                dp[i][j]=inf;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                mp[i][j]=read();
        dp[1][0]=0,dp[0][1]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if((i+j)%2)
                {
                    dp[i][j]=inf;
                    if(i>1&&j>1)
                        dp[i][j]=min(dp[i][j],dp[i-1][j-1]+min(mp[i-1][j]*mp[i][j],mp[i][j-1]*mp[i][j]));
                    if(i>1)
                        dp[i][j]=min(dp[i][j],dp[i-2][j]+mp[i-1][j]*mp[i][j]);
                    if(j>1)
                        dp[i][j]=min(dp[i][j],dp[i][j-2]+mp[i][j]*mp[i][j-1]);
                }
            }
        }
        printf("%d\n",dp[n][m]);
    }
}