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.
tips：广搜入门题目。感觉广搜比深搜容易理解一点，因为不用递归，递归太过于抽象，也难debug。
数轴上，要从n到k去，每次三种变换坐标的方法（n+1，n-1,n*2)
用队列储存n，然后出队n，进队三个新坐标，如此循环，每次进队都要判断新坐标和k是否相等，不相等就step[next] = step[head] + 1；相等则return step[next]。
#include <bits/stdc++.h>
using namespace std;
#define maxn 100001
queue<int>q;
int step[maxn];
int vis[maxn];
int bfs(int n, int k){
int head,next;
q.push(n);//进队
step[n] = 0;//步数初值为0
vis[n] = 1;//标记这个点已经走过了
while(!q.empty()){
head = q.front();//把队首（n)的值赋给head，这一步必须放在循环外面。不然三个坐标会尝试三次。
q.pop();//出队
for(int i = 0; i < 3; i++){
if(i == 0) next = head-1;
else if(i == 1) next = head+1;
else next = 2*head;//三个新坐标
if(next > maxn || next < 0) continue;//考虑越界
if(!vis[next]){
q.push(next);//next进队
step[next] = step[head] + 1;//步数+1，ps，这里用的step[head]+1，在每次循环的时候，他的初始值是一样的。
vis[next] = 1;//标记为旧点
}
if(k == next) return step[next];
}
}
}
int main(){
int n,k;
cin>>n>>k;
memset(vis,0,sizeof(vis));
memset(step,0,sizeof(step));
while(!q.empty()) q.pop();
if(n>k) cout<<n-k<<endl;
else cout<<bfs(n,k)<<endl;
return 0;
}