Catch That Cow

D - Catch That Cow

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

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?

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.

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<queue>#include<stack>#include<iostream>#include<algorithm>using namespace std;const int MAXN = 100010;int vis[MAXN];int n,k;struct node{int x,c;}cur,next;int bfs(){queue<node>q;while(!q.empty() )q.pop() ;memset(vis,0,sizeof(vis));//node cur,next;cur.x = n,cur.c = 0;vis[cur.x] = 1;q.push(cur);while(!q.empty() ){cur = q.front() ;q.pop() ;for(int i = 0; i < 3; i++){if(i == 0)next.x = cur.x-1;else if(i == 1)next.x = cur.x+1;elsenext.x = cur.x*2;next.c = cur.c+1;if(next.x == k)return next.c;if(next.x>=0 && next.x<=MAXN && !vis[next.x])            {                vis[next.x]=1;                q.push(next);            } }} return 0;}int main(){while(~scanf("%d%d",&n,&k)){if(n>=k){printf("%d\n",n-k);continue;}else{printf("%d\n",bfs());}}return 0;}