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.
分析：找最短路问题，首先想到BFS，三个状态分别为x-1，x+1，2*x，注意剪枝一些不太可能的状态，代码如下：
const int maxm = 1000100; struct Node { int times, x; Node(int _times, int _x):times(_times),x(_x) {} }; int vis[maxm], n, k; int main() { scanf("%d%d", &n, &k); queue<Node> q; q.push(Node(0, n)); while(!q.empty()) { Node tmp = q.front(); q.pop(); if(vis[tmp.x]) continue; vis[tmp.x] = 1; if(tmp.x == k) { printf("%d ", tmp.times); break; } tmp.times++; if(tmp.x && !vis[tmp.x-1]) q.push(Node(tmp.times, tmp.x - 1)); if(!vis[tmp.x+1]) q.push(Node(tmp.times, tmp.x + 1)); if(tmp.x < k * 4 && !vis[tmp.x*2]) q.push(Node(tmp.times, tmp.x * 2)); } return 0; }