FatMouse and Cheese

最新推荐文章于 2024-07-30 10:30:00 发布

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最新推荐文章于 2024-07-30 10:30:00 发布

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本篇介绍了一道经典的深搜算法题目——肥鼠吃奶酪问题。该问题要求计算肥鼠在遵循特定规则的情况下能吃到的最大奶酪数量。文章提供了完整的AC代码，并详细解释了使用记忆化搜索来解决此问题的方法。

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FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9844 Accepted Submission(s): 4156

Problem Description
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

解题思路：这道题是一道深搜题，还利用记忆化搜索，题意是肥老鼠要去吃洞里吃奶酪，每个洞口都有一个猫在守着，为了避免被猫抓住，它每次到洞口前最多跑k步，且每次吃的奶酪都要比上一次多，计算肥老鼠最多吃多少奶酪
AC代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int n,k;
int Map[1005][1005];
int dp[1005][1005];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int DFS(int x,int y)
{
    int ans=0;
    int max1=0;
    if(!dp[x][y])
    {
        for(int j=1;j<=k;j++)//每次吃奶酪前最多走k步
        {
            for(int i=0;i<4;i++)//上下左右四个方向
            {
                int xx=x+dir[i][0]*j;
                int yy=y+dir[i][1]*j;
                if(xx>=0&&xx<n&&yy>=0&&yy<n&&Map[x][y]<Map[xx][yy])
                {
                    ans=DFS(xx,yy);//满足条件，则再一次深搜
                    max1=max(max1,ans);//得到当前步由下一步得到的最多奶酪
                }
            }
        }
        dp[x][y]=max1+Map[x][y];//返回这一层得到的奶酪
    }
    return dp[x][y];
}
int main()
{
    int x,y,z;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(Map,0,sizeof(Map));
        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",&Map[i][j]);
        }
        cout<<DFS(0,0)<<endl;
    }
    return 0;
}