Catch That Cow

CatchThatCow算法解析

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 52 阅读

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#java

本文介绍了一个名为CatchThatCow的问题，这是一个经典的搜索问题。问题描述了农夫约翰如何使用步行和传送两种方式来抓住逃跑的奶牛。文章通过示例详细解释了如何使用广度优先搜索算法解决该问题，并提供了一段实现此算法的C++代码。

Catch That Cow

Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 113 Accepted Submission(s) : 46

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

4

 1 #include<stdio.h>
 2 #include<queue>
 3 #include<string.h>
 4 using namespace std;
 5 const int MAXN=100010;
 6 int visit[MAXN];
 7 int now[3];
 8 int N,K,step,t;
 9 void bfs(){
10     queue<int>dl;
11     dl.push(N);
12     visit[N]=1;
13     if(N!=K){
14     while(!dl.empty()){
15         t=dl.size();
16         step++;
17         while(t--){
18             now[0]=dl.front()+1;
19             now[1]=dl.front()-1;
20             now[2]=dl.front()*2;
21             dl.pop();
22             for(int i=0;i<3;i++){
23                 if(now[i]==K)return;
24                 if(now[i]>0&&now[i]<=100000&&!visit[now[i]]){//醉了，少了个0，错了n次；；；；； 
25                     visit[now[i]]=1;dl.push(now[i]);
26                 }
27             }
28         }
29     }
30 }
31 }
32 int main(){
33     while(~scanf("%d%d",&N,&K)){
34         memset(visit,0,sizeof(visit));
35         step=0;
36         bfs();
37         printf("%d\n",step);
38     }
39     return 0;
40 }