nyoj 234 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 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

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]
,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

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