Catch That Cow（BFS广搜）

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

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.

 

 

广搜的基本例题：

 

 #include<stdio.h>
 #include<string.h>
 #include<iostream>
 #include<queue>
 using namespace std;
 struct point
 {
     int x;///记录位置
     int count;///记录步数
 };
 queue<point>q;
 struct point s,now,t;
 int vis[];///假设FJ开始的位置就是100000，那么变化两倍之后就是200000
 int bfs(int n,int m)
 {
     int j;
     while(!q.empty())
     {
         q.pop();
     }///清空队列
     memset(vis,,sizeof(vis));
     vis[s.x]=;
     q.push(s);
     while(!q.empty())
     {
         t=q.front();
         if(t.x==m)
             return t.count;
         for(j=; j<; j++)
         {
             now=t;
             if(j==)
             {
                 now.x=now.x+;
             }
             else if (j==)
             {
                 now.x=now.x-;
             }
             else if(j==)
             {
                 now.x=now.x*;
             }
             now.count++;
             if(now.x==m)
             {
                 return now.count;
             }
             if(now.x>=&&now.x<=&&vis[now.x]==)
             {
                 vis[now.x]=;
                 q.push(now);
             }
         }
         q.pop();
     }
     return ;///二者开始的位置相同
 }
 int main()
 {
     int n,m,ans;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         s.x=n;
         s.count=;
         ans=bfs(n,m);
         printf("%d\n",ans);
     }
     return ;
 }

反思：这道题和之前的那一道剑客救公主那一道题一样，不仅仅需要考虑题意之中的搜索方式，还要考虑搜索不到或者起始位置与终止位置相同等特殊情况，该去如何设置被调函数，该去返回一个什么样的值，这两道题都是因为这一点使我wa了好多次，引以为戒。