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

6
0 7 5 13 9
4 81 6 22 4
40 9 34 16 10
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[200010], dp[] = new int[200010], DP[] = new int[200010], b[] = new int[200010];
    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 = 0; i < M; i++){
                DP[i] = 0;
                for(int j = 0; j < N; j++){
                    dp[j] = 0;
                    a[j] = cin.nextInt();
                    if(j < 2)
                        dp[j] = a[j];
                    else 
                        dp[j] = Math.max(dp[j - 2] + a[j], dp[j - 1]);
                }
                DP[i] = dp[N - 1];
                if(i < 2)
                    b[i] = DP[i];
                else
                    b[i] = Math.max(b[i - 2] + DP[i], b[i - 1]);
            }
            System.out.println(b[M - 1]);
        }
        
    }
}