hdoj 5569 matrix 【裸dp】

matrix

       Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 Total Submission(s): 35    Accepted Submission(s): 26

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 isa1∗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：给定n*m矩阵(保证n+m为奇数)，每个位置(i, j)有一个价值a[i][j]。现在要从(1, 1) 到达 (n, m)，每次只能向下或者向右走，依次记录途中经过位置a[1] ... a[2*k+1]。要求最后的a[1]*a[2] + ... + a[2*k]*a[2*k+1]值最小，问最小的值。

很裸的dp吧，用dp[i][j]记录到达位置(i, j)时胡最小值。然后暴力转移就好了。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define MAXN 100
#define MAXM 1010
#define eps 1e-8
#define LL long long
using namespace std;
LL a[1100][1100];
LL dp[1100][1100];
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                scanf("%lld", &a[i][j]);
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                if(i == 1 && j == 1)
                    dp[i][j] = 0;
                else if((i+j) & 1)
                {
                    if(i == 1)
                        dp[i][j] = dp[i][j-1] + a[i][j] * a[i][j-1];
                    else if(j == 1)
                        dp[i][j] = dp[i-1][j] + a[i][j] * a[i-1][j];
                    else
                        dp[i][j] = min(dp[i-1][j]+a[i][j]*a[i-1][j], dp[i][j-1]+a[i][j]*a[i][j-1]);
                }
                else
                {
                    if(i == 1)
                        dp[i][j] = dp[i][j-1];
                    else if(j == 1)
                        dp[i][j] = dp[i-1][j];
                    else
                        dp[i][j] = min(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        printf("%lld\n", dp[n][m]);
    }
    return 0;
}