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?
农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来．
他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有
两种办法移动，步行和瞬移：步行每秒种可以让约翰从z处走到x+l或x-l处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．
那么，约翰需要多少时间抓住那只牛呢？
Input
* Line 1: Two space-separated integers: N and K
仅有两个整数N和K.
Output
* Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
最短的时间．
Sample Input
5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.
Farmer John starts at point 5 and the fugitive cow is at point 17.
Sample Output
4
OUTPUT DETAILS:
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.
OUTPUT DETAILS:
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.
当$n<=1e5$时，它就是道简单的bfs。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 using namespace std; 6 #define N 100002 7 int vis[N<<1],s,t; 8 queue <int> h; 9 int main(){ 10 scanf("%d%d",&s,&t); h.push(s); 11 if(s==t){printf("0");return 0;} 12 while(!h.empty()){ 13 int x=h.front(); h.pop(); 14 if(!vis[x-1]&&x) vis[x-1]=vis[x]+1,h.push(x-1); 15 if(!vis[x+1]&&x+1<N) vis[x+1]=vis[x]+1,h.push(x+1); 16 if(!vis[x<<1]&&(x<<1)<N) vis[x<<1]=vis[x]+1,h.push(x<<1); 17 if(vis[t]){printf("%d",vis[t]);return 0;} 18 } 19 }