Beans

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

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 ?

Input

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<=200000.

Output

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

Sample Input

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

Sample Output

242

Source

2009 Multi-University Training Contest 4 - Host by HDU

Recommend

先每行求一下最大值，dp1[i] = max(dp[i - 1], dp[i - 2] + cow[i]);

再对这些最大值求一个最大值

dp2[i] = max(dp2[i - 1], dp2[i - 2] + col[i]);

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
int cow[200010];
int dp[200010];
int col[200010];
int dp2[200010];

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		memset (dp2, 0, sizeof(dp2));
		for (int i = 1; i <= n; ++i)
		{
			memset (dp, 0, sizeof(dp));
			for (int j = 1; j <= m; ++j)
			{
				scanf("%d", &cow[j]);
			}
			dp[1] = cow[1];
			for (int j = 2; j <= m; ++j)
			{
				dp[j] = max(dp[j - 1], dp[j - 2] + cow[j]);
			}
			col[i] = dp[m];
		}
		dp2[1] = col[1];
		for (int i = 2; i <= n; ++i)
		{
			dp2[i] = max(dp2[i - 1], dp2[i - 2] + col[i]);
		}
		printf("%d\n", dp2[n]);
	}
	return 0;
}