D - Catch That Cow BFS

The farmer knows the location of a cow, you want to grab it. And cattle farmers are in logarithmic axis, farmer starting at point N (0 <= N <= 100000), bovine positioned at the point K (0 <= K <= 100000). The farmer has two movements: 1, to move from the X or X-1 X + 1, 2 each take a minute movement, to move from X 2 * X, take a minute movement every cow is assumed unaware farmer action, not standing still. How much time it takes to seize at least cow? Input

Line: the two letters separated by a space:  N  and  K

Output

Line: cattle farmer grabbed minimum time required in minutes

Sample Input

5 17

Sample Output

4

Hint

Use the shortest possible time to seize the cattle farmer's programs are as follows: 5-10-9-18-17, 4 minutes.

// use Long Long 
#include <the iostream> 
#include <CString> 
#include <Queue>
 the using  namespace STD; 
typedef Long  Long LL; 
LL n-, K; 
const  int   N = 100001 ;
 int ARR [N] = { 0 } ; // record the number of steps
 int B [N] = { 0 }; // tag array
 int A [ . 3 ] = { . 1 , - . 1 , 2 };
 void the BFS ( int Start) { 
    Queue <LL> S;  
    S. push (start); // will start typing
    B [ Start] =0;//开始赋0
    arr[start]=1;标记
    while(s.size()){
        ll q=s.front();
        s.pop();

        if(q==k){
            break;
        }
        int x;
        for(int i=0;i<3;i++){
            if(a[i]==2){
                x=q*2;    
            }
            else {
                x=q+a[i];
            }
            if(X> = 0 && X <= N && ARR [X] == 0 ) { 
                B [X] = B [Q] + . 1 ; // add a new 
                s.push (X); 
                ARR [X] = . 1 ;
                 IF (x == k) reaches k // If x out
                     BREAK ; 
            } 
        } 
    } 
} 

int main () { 

    the while (CIN >> >> n- k) { 
        Memset (ARR, 0 , the sizeof (ARR)); 
        the BFS ( n-); 
        COUT << B [K] << endl; 
    } 
    
    
}