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.
题解：
BFS，注意剪枝。
#include<iostream> #include<cstring> #include<queue> using namespace std; const int maxn=1e5+5; bool vis[maxn*2]; struct node { int x,step; }; int k; int bfs(int n) { queue<node> que; node now; now.x=n; now.step=0; que.push(now); while(que.size()) { now=que.front(); que.pop(); node next; if(now.x==k) return now.step; if(now.x<k&&!vis[now.x+1]) { next.x=now.x+1; vis[next.x]=true; next.step=now.step+1; que.push(next); } if(now.x>0&&!vis[now.x-1]) { next.x=now.x-1; vis[next.x]=true; next.step=now.step+1; que.push(next); } if(now.x<k&&!vis[now.x*2]) { next.x=now.x*2; vis[next.x]=true; next.step=now.step+1; que.push(next); } } return -1; } int main() { int n; while(cin>>n>>k) { memset(vis,false,sizeof(vis)); cout<<bfs(n)<<endl; } return 0; }