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
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 3 int bfs(int n,int k); 4 struct node 5 { 6 int x,s; 7 }; 8 struct node q[400010]; 9 int book[100001];//标记数组，否则超内存 10 11 int main() 12 { 13 int n,k; 14 while(scanf("%d%d",&n,&k) != EOF) 15 { 16 if(n==k) 17 printf("0 "); 18 else 19 printf("%d ",bfs(n,k)); 20 } 21 return 0; 22 } 23 int bfs(int n,int k) 24 { 25 int i,head,tail,tx; 26 head=1; 27 tail=1; 28 q[tail].x=n; 29 q[tail].s=0; 30 book[n]=1; 31 tail++; 32 33 while(head < tail) 34 { 35 for(i=1;i<=3;i++) 36 { 37 if(i==1) 38 tx=q[head].x-1; 39 if(i==2) 40 tx=q[head].x+1; 41 if(i==3) 42 tx=q[head].x*2; 43 44 if(tx < 0 || tx > 100000) 45 continue; 46 if(!book[tx]) 47 { 48 book[tx]=1; 49 q[tail].x=tx; 50 q[tail].s=q[head].s+1; 51 tail++; 52 53 if(tx == k) 54 return q[tail-1].s; 55 } 56 } 57 head++; 58 } 59 }