吃土豆

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

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

样例输出

242

考虑对于某行某列元素，row[i][j]表示加上位置为i,j的土豆的质量的i行j列最大的和

列的最大值：row[i][j]=max(row[i][j-2]+row[i][j-3])+val

看图说话：

假设红色的格子为i行j列，那么它的前面有两种选择方案：

1、选择蓝色格子

2、选择黄色格子

那么该行最大的和是什么呢？

由于n列、n-1列具有状态无关性（n-1列的状态影响不了n列的状态），很显然等于max(row[i][n],row[i][n-1])

 

同理对于dp[i] (i行的最大值)

dp[i]=max(dp[i-2],dp[i-3])+max_row[i]

看图说话:

max土豆质量=max(dp[m],dp[m-1])

为了方便计算，我的代码把n，m扩大了2

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int e[505][505],dp[505];int n,m;int main(){while(scanf("%d%d",&m,&n)!=EOF){memset(e,0,sizeof(dp));memset(dp,0,sizeof(dp));int i,j,map;for(i=3;i<m+3;++i){for(j=3;j<n+3;++j){scanf("%d",&map);e[i][j]=max(e[i][j-2],e[i][j-3])+map;}}for(i=3;i<m+3;++i){dp[i]=max(dp[i-2],dp[i-3])+max(e[i][n+1],e[i][n+2]);}printf("%d\n",max(dp[m+1],dp[m+2]));}return 0;}