HDU 2845 Beans

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

分析：横竖分别求一下不连续的最大子段和;状态方程: Sum[i]=max(sum[j])+a[i];其中,0<=j<i-1; 

                                        or: sum[i]=max(sum[i-2]+a[i],sum[i-1])

code:

#include<stdio.h>
#define max(a,b)((a)>(b))?(a):(b)
#define N 200005   
int a[N],b[N],dp[N]; 
int main()  
{  
    int m,n;  
    int i,j;  
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
        for(i=1;i<=n;i++)  
        {  
            for(j=1;j<=m;j++)  
                scanf("%d",&a[j]);
            for(j=2;j<=m;j++)
                a[j]=max(a[j-1],a[j-2]+a[j]);
            b[i]=a[m]; 
        }  
        for(j=2;j<=n;j++)
            b[j]=max(b[j-1],b[j-2]+b[j]);  
        printf("%d\n",b[n]);  
    }  
    return 0;  
}