NYOJ 234 吃土豆(基础dp)

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

//横向dp，最后纵向dp注意最后的max

#include<iostream>
#include<algorithm>
#include<cstdio>

using namespace std;
int map[500+10][500+10]={0};
int main()
{
	int m,n;
	while(cin>>m>>n)
	{
		for(int i=1; i<=m; i++)
			for(int j=1; j<=n; j++)
				cin>>map[i][j];
		for(int i=1; i<=m; i++)
		{
			for(int j=3; j<=n; j++)
			{
				map[i][j]+=max(map[i][j-2],map[i][j-3]);
			}
			map[i][n]=max(map[i][n],map[i][n-1]);//注意是在两个间取最大
		}
		for(int i=3; i<=m; i++)
		{
			map[i][n]+=max(map[i-3][n],map[i-2][n]);
		}
		printf("%d\n",max(map[m][n],map[m-1][n]));
	}
	return 0;
}//ac