Beans(dp,两次dp)

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4141    Accepted Submission(s): 1964

Problem Description

Bean-eating is an interesting game, everyone owns an M*N matrix, which is filled with different qualities beans. Meantime, there is only one bean in any 1*1 grid. Now you want to eat the beans and collect the qualities, but everyone must obey by the following rules: if you eat the bean at the coordinate(x, y), you can’t eat the beans anyway at the coordinates listed (if exiting): (x, y-1), (x, y+1), and the both rows whose abscissas are x-1 and x+1.Now, how much qualities can you eat and then get ?

 

Input

There are a few cases. In each case, there are two integer M (row number) and N (column number). The next M lines each contain N integers, representing the qualities of the beans. We can make sure that the quality of bean isn't beyond 1000, and 1<=M*N<=200000.

 

Output

For each case, you just output the MAX qualities you can eat and then get.

 

Sample Input

4 6
11 0 7 5 13 9
78 4 81 6 22 4
1 40 9 34 16 10
11 22 0 33 39 6

 

Sample Output

242

 

题解：一次对行dp，一次对列dp；

java超时了。。。c就不会

代码：

import java.util.Scanner;
 
public class Beans {
    static int a[] = new int[], dp[] = new int[], DP[] = new int[], b[] = new int[];
    public static void main(String[] argvs){
        int M, N;
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()){
            M = cin.nextInt();
            N = cin.nextInt();
            for(int i = ; i < M; i++){
                DP[i] = ;
                for(int j = ; j < N; j++){
                    dp[j] = ;
                    a[j] = cin.nextInt();
                    if(j < )
                        dp[j] = a[j];
                    else
                        dp[j] = Math.max(dp[j - ] + a[j], dp[j - ]);
                }
                DP[i] = dp[N - ];
                if(i < )
                    b[i] = DP[i];
                else
                    b[i] = Math.max(b[i - ] + DP[i], b[i - ]);
            }
            System.out.println(b[M - ]);
        }
 
    }
}