POJ 3278 Catch That Cow（模板——BFS）

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
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\n");
18         else
19         printf("%d\n",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
=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 }