hdu2845—Beans（dp）

题目链接：传送门

Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4967    Accepted Submission(s): 2325

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

4 611 0 7 5 13 978 4 81 6 22 41 40 9 34 16 1011 22 0 33 39 6

 

Sample Output

242

 

解题思路：先对行dp获取每一行能获得的最大元素和，再对列dp，获取整个矩形中最大的元素和。行和列的动态转移方程都是dp[i] = max(dp[i-1],dp[i-2]+data[i])

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <string>#include <stack>#include <queue>using namespace std;typedef long long ll;typedef pair<int,int>PA;const int N = 120;const int M = 200119;const int INF = 0x3fffffff;int data[M],row[M],column[M];//先对行进行dp，再对列进行dp//column[i]:对某一行，在第i个位置能获得的最大sum//row[i]:对所有列，在第i列能获得的最大sumint main(){    int n,m;    while( ~scanf("%d%d",&n,&m) ){        for( int i = 2 ; i <= n+1 ; ++i ){            column[0] = column[1] = 0;            for( int j = 2 ; j <= m+1 ; ++j ){                scanf("%d",&data[j]);                column[j] = max( column[j-1] , column[j-2]+data[j] );            }            row[i] = column[m+1];        }        row[0] = row[1] = 0;        for( int i = 2 ; i <= n+1 ; ++i ){            row[i] = max( row[i-1] , row[i]+row[i-2] );        }        printf("%d\n",row[n+1]);    }    return 0;}