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

 

Sample Output

 

Source

#include<iostream>

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<algorithm>

#include<queue>

using namespace std;

const int N = ;

int a[N][N], dp[N][N];

int main()

{

    int n, m, i, j;

    while(scanf("%d%d", &n, &m)!=EOF)

    {

        for(i=; i<=n; i++)

        for(j=; j<=m; j++)

            scanf("%d", &a[i][j]);

        memset(dp, 0x3f3f3f3f, sizeof(dp));

        for(i=; i<=n; i++)

        for(j=; j<=m; j++)

        {

            if(i== && j==)

                dp[i][j] = ;

            else if((i+j)&)

            {

                dp[i][j] = min(dp[i-][j]+a[i-][j]*a[i][j], dp[i][j-]+a[i][j-]*a[i][j]);

            }

            else

                dp[i][j] = min(dp[i-][j], dp[i][j-]);

        }

        printf("%d\n", dp[n][m]);

    }

    return ;

}