HDU 2845 Beans

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1433    Accepted Submission(s): 741

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 6 11 0 7 5 13 9 78 4 81 6 22 4 1 40 9 34 16 10 11 22 0 33 39 6

 

Sample Output

242

 

Source

2009 Multi-University Training Contest 4 - Host by HDU

 

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#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[2003][2003];
int rc[2003];
int main()
{
    int n,m;
    int i,j;
    int Max;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
          for(j=1;j<=m;j++)
                scanf("%d",&dp[i][j]);

        for(i=1;i<=n;i++)
        {   rc[0]=-100000000;
            rc[1]=Max=dp[i][1];
            for(j=2;j<=m;j++)
             {
                rc[j]=max(max(rc[j-2]+dp[i][j],dp[i][j]),rc[j-2]);
                rc[j]=max(rc[j],rc[j-1]);//开始写的时候少了这个
                Max=Max>rc[j]?Max:rc[j];
             }
          dp[i][0]=Max;
        }
        rc[0]=-100000000;
        rc[1]=Max=dp[1][0];//方程写对后，rc[1]忘记赋值，又一次WA
        for(i=2;i<=n;i++)
        {
          rc[i]=max(max(rc[i-2]+dp[i][0],dp[i][0]),rc[i-2]);
           rc[i]=max(rc[i],rc[i-1]);
          Max=Max>rc[i]?Max:rc[i];
        }
        printf("%d\n",Max);
    }
    return 0;
}