HDU 2845 Beans

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Problem DescriptionBean-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 ? 
InputThere 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. 
OutputFor 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

 
Source2009 Multi-University Training Contest 4 - Host by HDU 每行來一次最大非連續子列。完了壓縮後最後再來一次=。=，不過要符合條件啦

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>using namespace std;const int maxn=200020;int n,m;int sum[maxn],a[maxn];int main(){   int n,m;   while(~scanf("%d%d",&n,&m))   {       for(int i=1;i<=n;i++)       {           for(int j=1;j<=m;j++)             scanf("%d",&a[j]);           for(int j=2;j<=m;j++)             a[j]=max(a[j-2]+a[j],a[j-1]);           sum[i]=a[m];       }       for(int i=2;i<=n;i++)          sum[i]=max(sum[i-2]+sum[i],sum[i-1]);       printf("%d\n",sum[n]);   }   return 0;}