问题描述：
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.
解题思路：
bfs，三个方向搜索，x=x+1,x=x-1,x=x*2，最先搜索到的就是用时最短的。
代码：
#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
using namespace std;
int n,k;
int a[]; struct node
{
int x;
int step;
}now,net; int bfs(int x)
{
queue<node> q;
now.x=x;
now.step=;
q.push(now);
while(q.size())
{
now=q.front();
q.pop();
//三种情况
if(now.x==k)return now.step;
net.x=now.x+;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x-;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
net.x=now.x*;
if(net.x>=&&net.x<=&&a[net.x]==)
{
a[net.x]=;
net.step=now.step+;
q.push(net);
}
}
return -;
} int main()
{
while(~scanf("%d%d",&n,&k))
{
memset(a,,sizeof(a));
a[n]=;
int ans=bfs(n);
printf("%d\n",ans);
}
return ;
}