HDOJ 2845 Beans

不能取相邻的行或列对行和列分别DP，【0】：没取的最大值，【1】：取的最大值

 

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1963    Accepted Submission(s): 993

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

 

#include <iostream>

#include <cstdio>

using namespace std;

const int maxn=222222;

int a[maxn],dp[maxn][2];

int N,M;

int main()

{

while(scanf("%d%d",&M,&N)!=EOF)

{

    for(int i=1;i<=M;i++)

    {

        for(int j=1;j<=N;j++)

        {

            int k;

            scanf("%d",&k);

            if(j<=1)

            {

                dp[j][1]=k;

            }

            else

            {

                dp[j][0]=max(dp[j-1][0],dp[j-1][1]);

                dp[j][1]=dp[j-1][0]+k;

            }

        }

        a=max(dp[N][0],dp[N][1]);

    }

    for(int i=1;i<=M;i++)

    {

        if(i==0||i==1)

        {

            dp[0]=0;

            dp[1]=a;

        }

        else

        {

            dp[0]=max(dp[i-1][0],dp[i-1][1]);

            dp[1]=dp[i-1][0]+a;

        }

    }

    int ans=max(dp[0],dp[1]);

    cout<<ans<<endl;

}

    return 0;

}