题目链接：
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.
题意描述：
输入人和牛在坐标轴上的位置
按照人寻找的三种走路方式，问最短抓到牛的时间。
解题思路：
搜索题，求最短时间，使用BFS更合适一点。另外需要注意：
走过的点是不需要重复走的，因为既然走过就证明牛不在这个点，所以使用book数组标记一下就不会出现超内存（扩展无用点）和超时啦。
AC代码：
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];//标记数组，否则超内存
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 }