hdu 5569 matrix dp

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=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m;

int mp[][];

int dp[][];

const int inf = 1e9+;

int main()

{

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

    {

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

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

                dp[i][j]=inf;

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

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

                mp[i][j]=read();

        dp[][]=,dp[][]=;

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

        {

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

            {

                if((i+j)%)

                {

                    dp[i][j]=inf;

                    if(i>&&j>)

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

                    if(i>)

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

                    if(j>)

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

                }

            }

        }

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

    }

}