Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36079		Accepted: 11123

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.

 

【题目来源】

【题目大意】

在x轴的正半轴上，一个人的出发点是N,一头牛在另一点K，人可以有两种操作：

1.步行：+1 or -1

2.传送：*2

问你通过最少的步数到达牛的位置，需要多少步。

 

【题目分析】

就是一个裸的广搜，使用队列实现。

 

#include
      
       #include
       
        #include
        
         using namespace std;int N,K;struct now{    int x;    int step;};bool vis[1000010];int BFS(int x){    int tx;    now ans,tp;    queue
         
           que;    ans.x=x,ans.step=0;    vis[x]=true;    que.push(ans);    while(!que.empty())    {        tp=que.front();        que.pop();        tx=tp.x+1;   //no.1        if(tx<0||tx>1000000);        else        {            if(!vis[tx])            {                vis[tx]=true;                ans.x=tx,ans.step=tp.step+1,que.push(ans);                if(ans.x==K) return ans.step;            }        }        tx=tp.x-1;    //no.2        if(tx<0||tx>1000000);        else        {            if(!vis[tx])            {                vis[tx]=true;                ans.x=tx,ans.step=tp.step+1,que.push(ans);                if(ans.x==K) return ans.step;            }        }        tx=tp.x*2;   //no.3        if(tx<0||tx>1000000);        else        {            if(!vis[tx])            {                vis[tx]=true;                ans.x=tx,ans.step=tp.step+1,que.push(ans);                if(ans.x==K) return ans.step;            }        }    }}int main(){    while(cin>>N>>K)    {        if(N==K)        {            cout<<"0"<
         
        
       
      

 

当然还可以剪枝一下，比如说：如果人的坐标大于牛的坐标，这时就只需要做-1这一步就可以了。