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
     
      #include
      
       #include
       
        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]);    }}