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.
大致题意：在0至100000的数轴上给定起点与目标点，可以进行左移右移和坐标翻倍的操作，时间花费均为1，输出最小的时间花费
其实就是从起点开始bfs，以0为下限，100000为上限，代码如下：
1 #include<stdio.h> 2 #include<iostream> 3 #include<string.h> 4 #include<algorithm> 5 #include<queue> 6 using namespace std; 7 int n,k,vis[100005]; 8 int move[2]={1,-1}; 9 struct node{ 10 int x,t; 11 }; 12 int bfs() 13 { 14 node now,next; 15 queue<node>q; 16 now.x=n,now.t=0,vis[n]=1; 17 q.push(now); 18 while(!q.empty()) 19 { 20 now=q.front(); 21 q.pop(); 22 if(now.x==k) return now.t; 23 for(int i=0;i<3;++i) 24 { 25 if(i==0||i==1) 26 { 27 next.x=now.x+move[i]; 28 if(next.x<0||next.x>100000||vis[next.x]) continue; 29 next.t=now.t+1,vis[next.x]=1; 30 q.push(next); 31 } 32 else 33 { 34 next.x=now.x*2; 35 if(next.x<0||next.x>100000||vis[next.x]) continue; 36 next.t=now.t+1,vis[next.x]=1; 37 q.push(next); 38 } 39 } 40 } 41 return 0; 42 } 43 int main() 44 { 45 scanf("%d%d",&n,&k); 46 printf("%d",bfs()); 47 return 0; 48 }