题目链接：http://i.cnblogs.com/EditPosts.aspx?opt=1
Catch That Cow
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7909 Accepted Submission(s):
2498
Problem 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.
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题目大意：在一个数轴上有n和k，农夫在n这个位置，奶牛在k那个位置，农夫要抓住奶牛，有两种方法：1、walking即农夫可以走x+1的位置和x-1的位置。
2、teleporting即每分钟可以走到2*x的位置。利用这两种方法，找到最快几步可以到达！
解题思路：其实就这三种情况，本来没打算用搜索的，不过发现好多东西都可以灵活运用！我们吧第一种方法看成两个方向，用dir[2]={1,-1,}表示，然后进行广搜！第二种方法自然就要特殊判断一下了！有一个地方要特殊判断一下，就是因为数据比较大，vis[ss.x]可能会超范围导致RE，所以都加一行判断使其保证在题目范围中。
详见代码。
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cmath>
5 #include <queue>
6
7 using namespace std;
8
9 int dir[3]= {1,-1};
10
11 struct node
12 {
13 int x,step;
14 } s,ss;
15
16 int bfs(int n,int k)
17 {
18 queue<node>q,qq;
19 s.x=n;
20 s.step=0;
21 int vis[100010]= {0};
22 q.push(s);
23 while (!q.empty())
24 {
25 s=q.front();
26 q.pop();
27 if (s.x==k)
28 return s.step;
29 for (int i=0; i<2; i++)
30 {
31 ss.x=s.x+dir[i];
32 ss.step=s.step+1;
33 if (ss.x>=0&&ss.x<=100000)
34 if (!vis[ss.x])
35 {
36 vis[ss.x]=1;
37 q.push(ss);
38 }
39 }
40 ss.x=s.x*2;
41 ss.step=s.step+1;
42 if (ss.x>=0&&ss.x<=100000)
43 {
44 if (!vis[ss.x])
45 {
46 vis[ss.x]=1;
47 q.push(ss);
48 }
49 }
50 }
51 return 0;
52 }
53
54 int main ()
55 {
56 int n,k;
57 while (~scanf("%d%d",&n,&k))
58 {
59 printf ("%d\n",bfs(n,k));
60 }
61 return 0;
62 }