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題目鏈接：
http://poj.org/problem?id=3278

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. 題意描述：
輸入人和牛在坐標軸上的位置 按照人尋找的三種走路方式，問最短抓到牛的時間。 解題思路： 搜索題，求最短時間，使用BFS更合適一點。另外需要註意： 　　走過的點是不需要重復走的，因為既然走過就證明牛不在這個點，所以使用book數組標記一下就不會出現超內存（擴展無用點）和超時啦。 AC代碼：

 1 #include<stdio.h>
 2 int bfs(int n,int k);
 3 struct node
 4 {
 5     int x,s;
 6  };
 7 struct node q[300010];
 8 int book[100001];//標記數組，否則超內存 
 9 int main()
10 {
11     int n,k;
12     while(scanf("%d%d",&n,&k) != EOF)
13     {
14         if(n==k)
15         printf("0\n");
16         else
17         printf("%d\n",bfs(n,k));
18     }
19     return 0;
20 }
21 int bfs(int n,int k)
22 {
23     int i,head,tail,tx;
24     head=1;
25     tail=1;
26     q[tail].x=n;
27     q[tail].s=0;
28     book[n]=1;
29     tail++;
30     while(head < tail)
31     {
32         for(i=1;i<=3;i++)
33         {
34             if(i==1)
35             tx=q[head].x-1;
36             if(i==2)
37             tx=q[head].x+1;
38             if(i==3)
39             tx=q[head].x*2;
40             if(tx < 0||tx > 100000)
41             continue;
42             if(!book[tx])
43             {
44                 book[tx]=1;
45                 q[tail].x=tx;
46                 q[tail].s=q[head].s+1;
47                 tail++;    
48             }
49             if(tx == k)
50             return q[tail-1].s;
51         }
52         head++;
53     }
54  } 

POJ 3278 Catch That Cow（模板——BFS）