NYOJ-234 吃土豆

 

吃土豆

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

描述

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 ?

 

输入

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<=500.

输出

For each case, you just output the MAX qualities you can eat and then get.

样例输入

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

样例输出

242

来源

2009 Multi-University Training Contest 4

 

题解：

仔细想想，其实这道题横坐标和纵坐标的状态是一样的，
对于取点(i, j)：
    则在列上，不能取旁边相邻的两列，即 j-1 列和 j+1 列；
    同理，在行上，不能取旁边相邻的两行，即 i-1 行和 i+1 行。
所以，可以对行和列单独考虑：
    对于列级状态转移方程： map[i][j] = max (map[i][j-2],map[i][j-3]);
    对于行级状态转移方程： dp[i] = max(dp[i-2], dp[i-3]) + max( map[i][n-1], map[i][n]);(假定一共有n列)

参考 ：http://www.cnblogs.com/A-way/archive/2013/05/08/3066236.html

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

int main()
{
    int n, m;
    int map[505][505], dp[505]; 
    
    while(~scanf("%d%d", &n, &m))
    {
        memset(map, 0, sizeof(map));
        memset(dp, 0, sizeof(dp));
        
        for(int i = 3; i < n+3; ++i)
            for(int j = 3; j < m+3; ++j)
            {
                scanf("%d", &map[i][j]);
                map[i][j] += max(map[i][j-2], map[i][j-3]);
            }
        
        for(int i = 3; i < n+3; ++i)
            dp[i] = max(dp[i-2], dp[i-3]) + max(map[i][m+1], map[i][m+2]);
        printf("%d\n", dp[n+2]);
    }
    return 0;
}