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[];
 int dp[][];
 int a[];
 int main()
 {
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         memset(x,,sizeof(x));
         for(int i=;i<=n;i++)
         {
             memset(dp,,sizeof(dp));
             for(int j=;j<=m;j++)
             {
                 scanf("%d",&a[j]);
             }
             dp[][]=a[];
             dp[][]=;
             for(int j=;j<=m;j++)
             {
                 dp[j][]=dp[j-][]+a[j];
                 dp[j][]=max(dp[j-][],dp[j-][]);
             }
             x[i]=max(dp[m][],dp[m][]);
         }
         memset(dp,,sizeof(dp));
         dp[][]=x[];
         dp[][]=;
         for(int i=;i<=n;i++)
         {
             dp[i][]=dp[i-][]+x[i];
             dp[i][]=max(dp[i-][],dp[i-][]);
         }
         int sum=max(dp[n][],dp[n][]);
         printf("%d\n",sum);
     }
     return ;
 }