hdoj problem 2845 Beans (动态规划)

Beans

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3002    Accepted Submission(s): 1446

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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/*
对于一行来说，相邻的数不可同时取，容易得到状态转移方程：
dp[i]=max(dp[i-2]+a[i],dp[i-1]);  
然后取每一行的最大得b[i] ,又可看作一行相同处理
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[200010];
int b[200010];
int dp[200010];
int main()
{
int  m,n;
while(~scanf("%d%d",&m,&n))
{
   int i,j,k;
   memset(dp,0,sizeof(dp));
   memset(b,0,sizeof(b));
  for(i=1;i<=m;i++)
  {
  for(j=1;j<=n;j++)
  scanf("%d",&a[j]);
 
       for(j=1;j<=n;j++)
         dp[j]=max(dp[j-2]+a[j],dp[j-1]);//此处不是dp[j]=max(dp[j-2],dp[j])
         b[i]=dp[n];
  }

   for(i=1;i<=m;i++)
     dp[i]=max(dp[i-2]+b[i],dp[i-1]);
     printf("%d\n",dp[m]);
}
return 0;
}