Catch That Cow（BFS）

最新推荐文章于 2025-09-13 09:06:35 发布

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最新推荐文章于 2025-09-13 09:06:35 发布

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文章标签： java

原文链接：http://www.cnblogs.com/a1225234/p/5036852.html

本文解析了一道名为“CatchThatCow”的算法题，该题目要求计算农夫约翰使用步行和传送两种方式追上静止不动的逃逸奶牛所需的最短时间。通过使用广度优先搜索算法（BFS），文章详细介绍了实现过程，并给出了解决方案。

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10166 Accepted Submission(s): 3179

Problem Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <queue>
 4 #include <cstring>
 5 using namespace std;
 6 int b,e;
 7 int Mintim;
 8 struct node
 9 {
10     int pos,t;
11 }k,tem;
12 int vis[100000+10];
13 queue<node> s;
14 void bfs()
15 {
16     while(!s.empty())
17         s.pop();
18     k.pos=b,k.t=0;
19     s.push(k);
20     while(!s.empty())
21     {
22         k=s.front();
23         s.pop();
24         if(k.pos==e)
25         {
26             Mintim=k.t;
27             return;
28         }
29         if(k.pos<0||k.pos>100000||vis[k.pos])    continue;
30         vis[k.pos]=1;
31         tem.t=k.t+1;
32         tem.pos=k.pos+1;
33         s.push(tem);
34         tem.pos=k.pos-1;
35         s.push(tem);
36         tem.pos=k.pos*2;
37         s.push(tem);
38     }
39 }
40 int main()
41 {
42     int i,j;
43     freopen("in.txt","r",stdin);
44     while(scanf("%d%d",&b,&e)!=EOF)
45     {
46         memset(vis,0,sizeof(vis));
47         bfs();
48         printf("%d\n",Mintim);
49     }
50     return 0;
51 }