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 * 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-), (x, y+), and the both rows whose abscissas are x- and x+.

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

样例输出

思路：

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

AC代码：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 using namespace std;

 #define N 506

 int n,m;

 int col[N][N];

 int dp[N];

 int main()

 {

     while(scanf("%d%d",&n,&m)==){

         memset(col,,sizeof(col));

         memset(dp,,sizeof(dp));

         for(int i=;i<n+;i++){

            for(int j=;j<m+;j++){

                int x;

                scanf("%d",&x);

                col[i][j] = max(col[i][j-],col[i][j-])+x;

            }

         }

         for(int i=;i<n+;i++){

             dp[i]=max(dp[i-],dp[i-])+max(col[i][m+],col[i][m+]);

         }

         printf("%d\n",max(dp[n+],dp[n+]));

     }

     return ;

 }