bfs find cow

#include<stdio.h>#include<string.h>#include<math.h>#include<stdlib.h>#include<algorithm>#include<iostream>#include<queue>using namespace std;struct move{int vis;int time;};int d[2] = { 1,-1 };int visa[200000];int bfs(int s,int e){if (s == e){return 0;}//int time = 0;struct move temp;struct move p;queue<struct move> q;memset(visa, 0, sizeof(visa));visa[s] = 1;p.vis = s;p.time = 0;q.push(p);while (!q.empty()){temp = q.front();q.pop();for (int i = 0; i < 3; i++){if (i == 0){p.vis = temp.vis + 1;}else if (i == 1){p.vis = temp.vis - 1;}else if (i == 2){p.vis = temp.vis * 2;}p.time = temp.time + 1;if (p.vis == e){return p.time;}if (p.vis >= 0 && p.vis < 200000 && !visa[p.vis]){visa[p.vis] = 1;q.push(p);}}}}int main(){int start, end;while (~scanf("%d %d", &start, &end)){printf("%d\n",bfs(start, end));}return 0;}//by tp

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.