hdu 2845 Beans

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2845

题目描述：

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

题目大意：

叫你求一个最大不连续子序列，要求横向不连续，纵向也不能连续，所以这里分别对横纵都求一下就好

Ac代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cstdlib>using namespace std;int a[200005],b[2000005];int m,n;int main(){    while(scanf("%d%d",&m,&n)!=EOF)    {        for(int i=1;i<=m;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&a[j]);            }            for(int j=2;j<=n;j++)            {                a[j]=max(a[j-2]+a[j],a[j-1]);            }            b[i]=a[n];        }        for(int i=2;i<=m;i++)        {            b[i]=max(b[i-2]+b[i],b[i-1]);        }         printf("%d\n",b[m]);    }    return 0;}