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.
题意：FJ要抓奶牛，输入n（FJ的位置）,k（奶牛的位置）,求FJ抓到奶牛所需最少时间。
FJ有三种走法：
1：向前移动一步，花费一分钟
2：向后退后一步，花费一分钟
3：向前移动当前位置的2倍，花费一分钟
1 #include <iostream> 2 #include <cstring> 3 #include <queue> 4 using namespace std; 5 const int MAX=100005; 6 bool vis[MAX]; 7 int step[MAX]; 8 queue <int >q; 9 int bfs(int n,int k) 10 { 11 int head,next; 12 q.push(n); 13 step[n]=0; 14 vis[n]=true; 15 while(!q.empty()) 16 { 17 head=q.front(); //取队首 18 q.pop(); 19 for(int i=0; i<3; i++) 20 { 21 if(i==0) 22 next=head-1; 23 if(i==1) 24 next=head+1; 25 if(i==2) 26 next=head*2; 27 if(next<0||next>=MAX) //排除出界情况 28 continue; 29 if(!vis[next]) 30 { 31 q.push(next); //入队 32 step[next]=step[head]+1;//步数加一 33 vis[next]=true; //标记访问 34 } 35 if(next==k) 36 return step[next]; 37 } 38 39 } 40 } 41 int main() 42 { 43 int n,k; 44 while(cin>>n>>k) 45 { 46 memset(vis,false,sizeof(vis)); 47 memset(step,0,sizeof(step)); 48 while(!q.empty()) 49 q.pop(); 50 if(n>=k) 51 cout<<n-k<<endl; 52 else 53 cout<<bfs(n,k)<<endl; 54 } 55 return 0; 56 }