hdu 1078 FatMouse and Cheese （DP）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1078

给出一个n*n的矩阵，每次最多走m步，而且下一步的值必须大于当前值，求最大值

从（0,0）位置开始递归，求最大值，最后dp[0][0]就是最大值

#pragma GCC optimize(2)
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;
typedef long long ll;
int n, m;
int mp[maxn][maxn];
int dp[maxn][maxn];
int d[4][2] = { 1,0,-1,0,0,1,0,-1 };
int dfs(int s, int t)
{
	int max = 0;
	if (dp[s][t] > 0)
	{
		return dp[s][t];
	}
	for (int i = 0; i < 4; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			int x1 = d[i][0] * j + s;
			int y1 = d[i][1] * j + t;
			if (x1<0 || y1<0 || x1>=n || y1>=n)
			{
				continue;
			}
			if (mp[s][t] < mp[x1][y1])
			{
				int tmp = dfs(x1, y1);
				if (tmp > max)
				{
					max = tmp;
				}
			}
		}
	}
	dp[s][t] = max + mp[s][t];
	return dp[s][t];
}
int main()
{
	//freopen("C://input.txt", "r", stdin);
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (n == -1 && m == -1)
		{
			break;
		}
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				scanf("%d", &mp[i][j]);
			}
		}
		memset(dp, 0, sizeof(dp));
		printf("%d\n", dfs(0, 0));
	}
	return 0;
}