HDU 1078 FatMouse and Cheese（记忆化搜索，DP）

╰+哭是因爲堅強的太久メ 2021-08-29 00:50 330阅读 0赞

### 目录 ###

*   *   *  1.题目
         *  2.题意
         *  3.思路
         *  4.代码

### 1.题目 ###

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

3 1
    1 2 5
    10 11 6
    12 12 7
    -1 -1

Sample Output  
37

### 2.题意 ###

老鼠每次只能走k步停下来，停下的这个位置只能比上一个停留的位置大，并获取其价值，每次只能水平或垂直走，问最大能得到的价值在n\*n的格子上，每个点各有若干块奶酪，胖老鼠从左上角出发，每次最多走k步（只能直走），且下一点必须比这一点的奶酪多，问最多能吃到多少块奶酪。

### 3.思路 ###

记忆化搜索  
虽然本质上是DFS这种搜索的思路，但其对搜索过的状态进行记录，从而完成对未知状态的推导，实际上也是一种DP的思想。

链接: [记忆化搜索专题][Link 1].

### 4.代码 ###

#define _CRT_SECURE_NO_WARNINGS
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    const int maxn = 100 + 5;
    int a[maxn][maxn], dp[maxn][maxn];  //dp[i][j]表示以第i行第j列个网格为起点所能得到的最大值
    int dirt[4][2] = {  { 0,1},{ 0,-1},{ 1,0},{ -1,0} };  ///四个方向
    int n, k;
    bool in(int x, int y) 判断是否出边界
    { 
    	if (x >= 0 && x < n&&y >= 0 && y < n)
    		return 1;
    	else
    		return 0;
    }
    int dfs(int x, int y)
    { 
    	int ans=0;
    	for (int j = 1; j <= k; j++)  ///水平或者垂直方向走k步
    	{ 
    		for (int i = 0; i < 4; i++)
    		{ 
    			int nx = x + dirt[i][0]*j;  ///水平
    			int ny = y + dirt[i][1]*j;  ///垂直
    			if (!in(nx, ny)) continue;
    			if (a[x][y] >= a[nx][ny])  continue;   //满足递增序列
    			if (dp[nx][ny])  //如果已经探索过了，就没必要继续深搜了
    			{ 
    				ans = max(ans, dp[nx][ny]); continue;
    			}
    			ans = max(ans, dfs(nx,ny));   //取所有的最大值
    		}
    	}
    	return dp[x][y] = a[x][y] + ans;
    }
    int main()
    { 
    	while(scanf("%d%d", &n, &k) != EOF)
    	{ 
    		memset(dp, 0, sizeof(dp));
    		if (n == -1 && k == -1)  break;
    		for (int i = 0; i < n; i++)
    			for (int j = 0; j < n; j++)
    				scanf("%d", &a[i][j]);
    		printf("%d\n", dfs(0, 0));  //最优解在出发点上
    	}
    	return 0;
    }

参考博客：  
链接: [点击这里][Link 2].

[Link 1]: https://blog.csdn.net/hjf1201/article/details/78680814/
[Link 2]: https://blog.csdn.net/computer_user/article/details/77089670/