Catch That Cow（广搜）

个人心得：其实有关搜素或者地图啥的都可以用广搜，但要注意标志物不然会变得很复杂，想这题，忘记了标志，结果内存超时；

将每个动作扔入队列，但要注意如何更简便，更节省时间，空间

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?

InputLine 1: Two space-separated integers: N and KOutputLine 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<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 using namespace std;
 int sum;
 int ok=;
 struct Node
 {
     int x;
     int y;

 };
 int book[];

 void dfs(int n,int m)
 {
     memset(book,,sizeof(book));
     queue<Node >s;
     book[n]=;
     Node t;
     t.x=n;t.y=;
     s.push(t);
     int a;
     Node tt;
     while(!s.empty())
     {
           a=s.front().x*;
         tt.x=a,tt.y=s.front().y+;
         if(a==m)
         {
             sum=tt.y;
             return ;

         }
          if(tt.x>=&&tt.x<=)
             if(!book[a])
            {
                book[a]=;
                s.push(tt);

                }
         a=s.front().x+;
         tt.x=a;
         tt.y=s.front().y+;
         if(a==m)
         {
             sum=tt.y;
            return ;

         }
       if(tt.x>=&&tt.x<=)
             if(!book[a])
            {
                book[a]=;
                s.push(tt);

                };
          a=s.front().x-;
         tt.x=a,tt.y=s.front().y+;
         if(a==m)
         {
             sum=tt.y;
             return ;

         }
         if(tt.x>=&&tt.x<=)
             if(!book[a])
            {
                book[a]=;
                s.push(tt);

                }
         s.pop();

     }
     return ;

 }
 int main()
 {

     int n,m;
     while(cin>>n>>m)
     {
         sum=;
         if(n>=m) sum=n-m;
        else
         dfs(n,m);
     cout<<sum<<endl;

     }
     return ;

 }