学步园

Catch That Cow

Time Limit:2000MS  Memory Limit:65536K
Total Submit:118 Accepted:30

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.

Source

#include<iostream><br /> #include<queue><br /> using namespace std;<br /> #define M 100001<br /> #define INF 10000000</p> <p>struct POINT<br /> {<br /> int pos;<br /> int step;<br /> }now,next;<br /> queue<POINT>Q;<br /> bool visited[M];</p> <p>int n,k;<br /> int bfs()<br /> {<br /> while(!Q.empty())<br /> Q.pop();<br /> now.pos=n;<br /> now.step=0;<br /> visited[now.pos]=true;<br /> Q.push(now);<br /> while(!Q.empty())<br /> {<br /> now=Q.front();<br /> Q.pop();<br /> next=now;<br /> for(int i=0;i<3;i++)<br /> {<br /> if(i==0)<br /> next.pos=now.pos+1;<br /> if(i==1)<br /> next.pos=now.pos-1;<br /> if(i==2)<br /> next.pos=now.pos*2;<br /> next.step=now.step+1;<br /> if(next.pos==k)<br /> return next.step;<br /> if(next.pos<0||next.pos>M)<br /> continue;<br /> if(!visited[next.pos])<br /> {<br /> visited[next.pos]=true;<br /> Q.push(next);<br /> }<br /> }<br /> }<br /> return INF;<br /> }</p> <p>int main()<br /> {<br /> while(cin>>n>>k)<br /> {<br /> memset(visited,false,sizeof(visited));<br /> if(n<k)<br /> cout<<bfs()<<endl;<br /> if(n==k)<br /> cout<<0<<endl;<br /> if(n>k)<br /> cout<<n-k<<endl;<br /> }<br /> }