Problem 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.
程序代码：
#include<iostream> #include<queue> #include<cstring> using namespace std; #define maxn 100001 bool vis[maxn]; int step[maxn]; queue<int> q;//定义队列 int bfs(int n,int k) { int head,next; q.push(n);//n入队 step[n]=0; vis[n]=true; while(!q.empty())//当队列非空 { head=q.front();//取队首 q.pop();//出列 for(int i=0;i<3;i++) { if(i==0) next=head-1; else if(i==1) next=head+1; else next=head*2; if(next<0||next>=maxn) continue; if(!vis[next]) { q.push(next); step[next]=step[head]+1; vis[next]=true; } if(next==k) return step[next]; } } } int main() { int n,k; while(cin>>n>>k) { memset(step,0,sizeof(step)); memset(vis,false,sizeof(vis)); while(!q.empty()) q.pop();//调用前先清空 if(n>=k) cout<<n-k<<endl; else cout<<bfs(n,k)<<endl; } return 0; }
