HDU - 1078-day8E

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本文探讨了FatMouse在一个n*n的城市网格中收集奶酪的问题，每次移动必须到达奶酪数量超过当前位置的地点，且移动距离受超级猫TopKiller的限制。通过深度优先搜索算法（DFS），求解FatMouse在无法移动前能收集到的最大奶酪数量。

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问题来源

HDU - 1078

问题描述

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he’s going to enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse – after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.

输入

There are several test cases. Each test case consists of

a line containing two integers between 1 and 100: n and k
n lines, each with n numbers: the first line contains the number of blocks of cheese at locations (0,0) (0,1) … (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), … (1,n-1), and so on.
The input ends with a pair of -1’s.

输出

For each test case output in a line the single integer giving the number of blocks of cheese collected.

例子

input

output

解题思路

dfs题，老鼠要吃奶酪，但下一次吃的奶酪要比这一次的多，并且可以使用冲刺，最多一次冲刺k格，问最多吃多少奶酪。
每一格都找到最多的奶酪的冲刺方法。叠加起来放入对应的sum数组里最后在起点（0，0）得到的是最多的奶酪数。

AC的代码

#include<cstring>
#include <iostream>
using namespace std;
int n, k;
int aa[105][105],sum[105][105];
int dx[] = { 0,-1,1,0 };
int dy[] = { -1,0,0,1 };
int dfs(int x, int y)
{
	int s=0;
	if (sum[x][y])
	{
		return sum[x][y];
	}
	for (int j = 0; j < 4; j++)
	{
		for (int p = 1; p <= k; p++)
		{
			int xx = x + dx[j] * p;
			int yy = y + dy[j] * p;
			if (xx >= 0 && xx < n&&yy >= 0 && yy < n&&aa[x][y] < aa[xx][yy])
			{
				int f1 = dfs(xx, yy);
				if (s < f1)
					s = f1;
			}
		}
	}

	sum[x][y] = s + aa[x][y];
	return sum[x][y];



}
int main()
{
   

	while (cin >> n >> k && (n != -1 && k != -1))
	{
		memset(sum, 0, sizeof(sum));
		for (int g = 0; g < n; g++)
		{
			for (int h = 0; h < n; h++)
			{
				scanf_s("%d", &aa[g][h]);
			}
		}
		dfs(0, 0);
		cout << sum[0][0]<<endl;
	}
	

}