hdu 2845 Beans

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3682    Accepted Submission(s): 1762

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

 

 

Recommend

gaojie   |   We have carefully selected several similar problems for you:  2830 2577 2870 2844 1069 

 

一道很明显的dp题，隔行隔列相加之和最大，同样的dp列式使用两次。

 

题意：取其中一个数，则不能取相邻的两个数，最后取出数字之和为每行的最大子序列之和。

再在每行的子序列之和中取出几个数字，规则和取行数字相同，不能取相邻的数，最后之和为输出最终结果。

 

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#define N 200005
using namespace std;
int a[N],b[N],dp[N];
int main()
{
    int m,n,i,j;
    while(~scanf("%d%d",&n,&m))
    {
        dp[0]=a[0]=b[0]=0;    //初始化
        for(i=1; i<=n; i++)
        {
            for(j=1; j<=m; j++)
                scanf("%d",&a[j]);
            dp[1]=a[1];
            for(j=2; j<=m; j++)
                dp[j]=max(dp[j-2]+a[j],dp[j-1]);   //dp思想，取结果大的数据
            b[i]=dp[m];     //b数组记录每一行的最大子序列和
        }
        dp[1]=b[1];
        for(i=2; i<=n; i++)
            dp[i]=max(dp[i-2]+b[i],dp[i-1]);   //完全一样的思想，只是前面去行的最大值，这里取列的最大值
        printf("%d\n",dp[n]);
    }
    return 0;
}