Beans(dp,两次dp)

时间:2023-03-09 06:22:36
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.Beans(dp,两次dp)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 - ]);
} }
}