matrix

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

Total Submission(s): 325    Accepted Submission(s): 196

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

 

 

Source

 

 写完这题， 我意识到了动态规划最重要的便是状态转移， 虽然以前听别人说， 但是到底没有自己真正体会到的来的贴切

 

 

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;const int N = 1100;int a[N][N], dp[N][N];int main(){ int n, m, i, j; while(scanf("%d%d", &n, &m)!=EOF) { for(i=1; i<=n; i++) for(j=1; j<=m; j++) scanf("%d", &a[i][j]); memset(dp, 0x3f3f3f3f, sizeof(dp)); for(i=1; i<=n; i++) for(j=1; j<=m; j++) { if(i==1 && j==1) dp[i][j] = 0; else 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]); } printf("%d\n", dp[n][m]); } return 0;}