Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 54911		Accepted: 17176

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.

题解：

宽度优先搜索bfs

代码：

#include
      
       #include
       
        #include
        
         using namespace std;#define N 100005struct Node{    int x;    int step;};queue
         
          q;bool visit[N];bool ok(int x){    if(x>=0 && x<=100000)        return 1;    return 0;}int cal(int i,int x){    if(i==0)        return x+1;    else if(i==1)        return x-1;    else        return x*2;}void bfs(int a,int b){    memset(visit,0,sizeof(visit));    while(!q.empty())    {        q.pop();    }    Node aa;    aa.step=1;    aa.x=a;    q.push(aa);    visit[aa.x]=1;    while(!q.empty())    {        Node tmp=q.front();        q.pop();        for(int i=0; i<3; i++)        {            int y=cal(i,tmp.x);            if(ok(y)&&!visit[y])            {                if(y==b)                    cout<
          
           <
           
            >n>>k; if(k>n) bfs(n,k); else cout<<(n-k); return 0;}