杭电2845 Beans 不连续的最大字段和

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3999    Accepted Submission(s): 1903

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

 

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看图意思就是求，横向的最大不连续子段和，然后得到一个数组，求这个数组的最大不连续子段和，就是相求横向然后求纵向：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[5500],i,j,k,l,m,n,s[5500],ans[5500];int main(){while(scanf("%d%d",&n,&m)==2){int help=0;int x=0;k=0; memset(ans,0,sizeof(ans));for(i=1;i<=n;i++){int cmt=0;int cnt=0;memset(a,0,sizeof(a));for(j=1;j<=m;j++){scanf("%d",&a[j]);a[j]+=cmt; cnt=max(cnt,a[j]);//横向不连续的最大子段和 cmt=max(cmt,a[j-1]);}ans[i]=cnt+x;k=max(k,ans[i]);//纵向不连续的最大子段和 x=max(x,ans[i-1]);}printf("%d\n",k);}}