hdu2845 Beans（DP）

思路：二维动态规划，先求每一行的最大和，组成新的一行，最后再求这一行的最大和

#include<stdio.h>
#include<string.h>
int r[200010][2],c[200010][2];
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
main()
{
int i,j,R,C,s;
while(scanf("%d%d",&R,&C)!=EOF)
{
memset(c,0,sizeof(c));
for(i=1;i<=R;i++)
{
for(j=1;j<=C;j++)
{
scanf("%d",&s);
r[j][0]=max(r[j-1][1],r[j-1][0]);
r[j][1]=r[j-1][0]+s;
}
c[i][1]=c[i-1][0]+max(r[j-1][1],r[j-1][0]);
c[i][0]=max(c[i-1][1],c[i-1][0]);
}
printf("%d\n",max(c[i-1][1],c[i-1][0]));
}
}

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