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#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <queue> using namespace std; int dir[3]= {1,-1}; struct node { int x,step; } s,ss; int bfs(int n,int k) { queue<node>q,qq; s.x=n; s.step=0; int vis[100010]= {0}; q.push(s); while (!q.empty()) { s=q.front(); q.pop(); if (s.x==k) return s.step; for (int i=0; i<2; i++) { ss.x=s.x+dir[i]; ss.step=s.step+1; if (ss.x>=0&&ss.x<=100000) if (!vis[ss.x]) { vis[ss.x]=1; q.push(ss); } } ss.x=s.x*2; ss.step=s.step+1; if (ss.x>=0&&ss.x<=100000) { if (!vis[ss.x]) { vis[ss.x]=1; q.push(ss); } } } return 0; } int main () { int n,k; while (~scanf("%d%d",&n,&k)) { printf ("%d\n",bfs(n,k)); } return 0; }
