HDU 5569 （BC#63）DP

matrix
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 143 Accepted Submission(s): 97

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
题意：
有一个矩阵，从（1，1）走到（n,m）使得a1∗a2+a3∗a4+…+a2k−1∗a2k.最小。
题解：
典型的DP，dp[i][j]为从（1，1）到（i，j）得到的最小权重，那么可以知道。可能是从（i-1，j）过来的，也可能是从（i，j-1）过来的。由于对于（i-1，j）和（i，j-1）也可能是从这两个方向过来的。所以有四种状态，那么每次我们选一个最小的递推就可以了。 。当时比赛做这个题的时候脑子发热。 。结果数组越界没看出来。 。wa了9次。 。现在在hdu上AC了。现把AC代码贴出来。

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#define F(i,a,b) for(int i = a;i<=b;i++)#define FI(i,a,b) for(int i = a;i>=b;i--)using namespace std;int a[1010][1010];int dp[1010][1010];int main(){//    freopen("data.in","r",stdin);//    freopen("data.out","w",stdout);    int m,n;    while(scanf("%d%d",&m,&n)!=EOF){        F(i,2,m+1)            F(j,2,n+1)                scanf("%d",&a[i][j]);        memset(dp,0x3f,sizeof(dp));        for(int i = 3;i<=n+1;i+=2)            if(i==3)                dp[2][i] = a[2][i]*a[2][i-1];            else                dp[2][i] = a[2][i]*a[2][i-1] + dp[2][i-2];        for(int i = 3;i<=m+1;i+=2)            if(i==3)                dp[i][2] = a[i][2]*a[i-1][2];            else                dp[i][2] = a[i][2]*a[i-1][2] + dp[i-2][2];        F(i,3,m+1){            for(int j = (i)%2+3;j<=n+1;j+=2){                int tem1 = a[i-1][j]*a[i][j]+dp[i-2][j];                int tem2 = a[i-1][j]*a[i][j]+dp[i-1][j-1];                int tem3 = a[i][j-1]*a[i][j]+dp[i-1][j-1];                int tem4 = a[i][j-1]*a[i][j]+dp[i][j-2];                dp[i][j] = min(min(tem1,tem2),min(tem3,tem4));            }        }        printf("%d\n",dp[m+1][n+1]);    }    return 0;}