HDU 2845 Beans

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5350    Accepted Submission(s): 2469

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
#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int i,j;
        int row[10002],dp[10002];
        memset(row,0,sizeof(row));
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
        {
            int c[10002];
            memset(c,0,sizeof(c));
            //c[j]记录每一行到第j个时所能获得的最大值
            int x;scanf("%d",&x);
            c[1]=x;
            for(j=2;j<=m;j++)
            {
                scanf("%d",&x);
                c[j]=max(c[j-1],c[j-2]+x);
            }
            row[i]=c[m];
        }
        dp[1]=row[1];
        int ma=dp[1];
        for(i=2; i<=n; i++)
        {
            dp[i]=max(dp[i-1],dp[i-2]+row[i]);
            ma=max(ma,dp[i]);
        }
        printf("%d\n",ma);
    }
    return 0;
}