HDU2845 DP

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4451    Accepted Submission(s): 2103

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

 

题意：

每一个格子有一个豆子，豆子有质量，每一行相邻的两个豆子不能同时选必须相隔，每一列也不能同时选，选了一列的某几个豆子则它的上一列和下一列不能选，问最多选到的豆子质量。

代码：

/*
两次dp,算出每一行的最大值，再用每一行的最大值组成一列算出这列的最大值即可。
*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int n,m;
int x[200000];
int dp[200000][2];
int a[200000];
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(x,0,sizeof(x));
        for(int i=1;i<=n;i++)
        {
            memset(dp,0,sizeof(dp));
            for(int j=1;j<=m;j++)
            {
                scanf("%d",&a[j]);
            }
            dp[1][1]=a[1];
            dp[1][0]=0;
            for(int j=2;j<=m;j++)
            {
                dp[j][1]=dp[j-1][0]+a[j];
                dp[j][0]=max(dp[j-1][0],dp[j-1][1]);
            }
            x[i]=max(dp[m][0],dp[m][1]);
        }
        memset(dp,0,sizeof(dp));
        dp[1][1]=x[1];
        dp[1][0]=0;
        for(int i=2;i<=n;i++)
        {
            dp[i][1]=dp[i-1][0]+x[i];
            dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
        }
        int sum=max(dp[n][1],dp[n][0]);
        printf("%d
",sum);
    }
    return 0;
}