hdu-5569 matrix(dp)

题目链接：

matrix

Time Limit: 6000/3000 MS (Java/Others)  

  Memory Limit: 65536/65536 K (Java/Others)

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

 

题意：

 

问怎么走才能使这个式子的值尽量小;

 

思路：

 

dp水题;

 

AC代码;

 

/************************************************ 
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┆┗━┓　  　┏━┛ ┆ 
┆　　┃　  　┃　　┆　　　　　　 
┆　　┃　  　┗━━━┓ ┆ 
┆　　┃　 AC代马 　　┣┓┆ 
┆　　┃　        　　┏┛┆ 
┆　　┗┓┓┏━┳┓┏┛ ┆ 
┆　　　┃┫┫　┃┫┫ ┆ 
┆　　　┗┻┛　┗┻┛ ┆       
************************************************ */  


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=1e3+4;
const double eps=1e-8;

int a[maxn][maxn],dp[maxn][maxn];

int main()
{       
        int n,m;
        while(cin>>n>>m)
        {
            For(i,1,n)For(j,1,m)read(a[i][j]);
            For(i,0,n)For(j,0,m)dp[i][j]=inf;
            dp[0][1]=dp[1][0]=0;
            For(i,1,n)
            {
                For(j,1,m)
                {
                    if((i+j)&1)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]);
                }
            }
            cout<<dp[n][m]<<"\n";
        }
        return 0;
}