HDU 2845 Beans

Beans

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 4

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

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int disx[110000];//行的最大值 
int disy[110000];//列的最大值 
int main()
{
	int n,m,i,j,k,l;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(disx,0,sizeof(disx));
		memset(disy,0,sizeof(disy));
		for(i=2;i<=n+1;i++)
		{
			for(j=2;j<=m+1;j++)
			{
				scanf("%d",&k);
				disy[j]=max(disy[j-1],disy[j-2]+k);
			}
			disx[i]=max(disx[i-1],disx[i-2]+disy[m+1]);
		}
		printf("%d\n",disx[n+1]);
	}
	return 0;
}