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

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

 

Sample Output

242

 

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

代码：

#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][0]表示第j列的值不选,则第j-1列可选可不选r[j][1]=r[j-1][0]+s;                     //第j列选，则前一列不选，加上第j列的值}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]));}}