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

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：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 #include<queue>
 5 using namespace std;
 6 struct point
 7 {
 8     int x;///记录位置
 9     int count;///记录步数
10 };
11 queue<point>q;
12 struct point s,now,t;
13 int vis[200000];///假设FJ开始的位置就是100000，那么变化两倍之后就是200000
14 int bfs(int n,int m)
15 {
16     int j;
17     while(!q.empty())
18     {
19         q.pop();
20     }///清空队列
21     memset(vis,0,sizeof(vis));
22     vis[s.x]=1;
23     q.push(s);
24     while(!q.empty())
25     {
26         t=q.front();
27         if(t.x==m)
28             return t.count;
29         for(j=0; j<3; j++)
30         {
31             now=t;
32             if(j==0)
33             {
34                 now.x=now.x+1;
35             }
36             else if (j==1)
37             {
38                 now.x=now.x-1;
39             }
40             else if(j==2)
41             {
42                 now.x=now.x*2;
43             }
44             now.count++;
45             if(now.x==m)
46             {
47                 return now.count;
48             }
49             if(now.x>=0&&now.x<=200000&&vis[now.x]==0)
50             {
51                 vis[now.x]=1;
52                 q.push(now);
53             }
54         }
55         q.pop();
56     }
57     return 0;///二者开始的位置相同
58 }
59 int main()
60 {
61     int n,m,ans;
62     while(scanf("%d%d",&n,&m)!=EOF)
63     {
64         s.x=n;
65         s.count=0;
66         ans=bfs(n,m);
67         printf("%d
",ans);
68     }
69     return 0;
70 }

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。