HDU 2845 Beans（dp）

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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

 

Sample Output

242

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

 

思路：注意状态转移方程，还有就是行的转移方程和列的相似，还有注意的是数组开大点

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 200005int dpx[N],dpy[N];int main(){    int n,m,i,j,s;    while(~scanf("%d%d",&n,&m))    {        memset(dpx,0,sizeof(dpx));        memset(dpy,0,sizeof(dpy));        for(i=2;i<=n+1;i++)        {             //memset(dpy,0,sizeof(dpy));             //这里为什么可以注释掉呢，因为会被覆盖                          for(j=2;j<=m+1;j++)            {                scanf("%d",&s);                dpy[j]=max(dpy[j-1],dpy[j-2]+s);  //dpy表示这一行从左到右能取到的最大的数和            }            dpx[i]=max(dpx[i-1],dpx[i-2]+dpy[1+m]);  //dpx表示从以上行中能取到数的最大的和        }        printf("%d\n",dpx[n+1]);    }    return 0;}