终于期末考试结束了，可以说，后期的速成真的很累，考的也不是很理想，下次在写一个反思贴吧，确实，，这学期让我学到了很多的东西．．．．
Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 49711 | Accepted: 15586 |
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.
Source
解题思路：
这道题其实就是一个一维的广搜问题，然后再加上一步求解最短路，记得以前在小优的博客上看到过，求最短路的搜索问题统一使用bfs较为简单，所以，这也算是一个３方向的搜索问题了，不像dfs一样，他需要回溯，他只需要状态扩展就可以了，所以，设置一个ｆｏｒ,写出ｉ的三种状态，然后对i的每种状态进行更新，next = head - 1; next = head + 1; next = head * 2, 对于step［ＭＡＸ]的每次的更新，一定要学会．．．好了，没什么说了，继续开搜索专题．
代码：
# include<cstdio>
# include<iostream>
# include<queue>
using namespace std;
# define MAX 100006
queue<int>x;
bool vis[MAX];
int step[MAX];
int n,m;
int bfs( int n,int k )
{
int head;
int next;
x.push(n);
vis[n] = true;
step[n] = 0;
while ( !x.empty() )
{
head = x.front();
x.pop();
for ( int i = 0;i < 3;i++ )
{
if ( i == 0 )
{
next = head - 1;
}
if ( i == 1 )
{
next = head + 1;
}
if ( i == 2 )
{
next = head * 2;
}
if ( next > MAX||next < 0 )
continue;
if ( !vis[next] )
{
x.push(next);
step[next] = step[head]+1;
vis[next] = true;
}
if ( next == k )
return step[next];
}
}
}
int main(void)
{
int n,m;
while ( cin>>n>>m )
{
if ( n >= m )
cout<<n-m<<endl;
else
cout<<bfs(n,m)<<endl;
}
return 0;
}