Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 118888		Accepted: 37099

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.

 

搜索入门，有坑点

#include 
    
     #include 
     
      #include 
      
       #define MAX 100001using namespace std;queue 
       
         Q;int dist[MAX];int BFS ( int n, int k ){    Q.push( n );    fill( dist, dist+MAX, 0);    dist[n] = 1;    while( !Q.empty() )    {        int v = Q.front();        Q.pop();        if( v == k )            break;        if( 2*v <= MAX && !dist[2*v] )  //坑点：不能写成  !dist[2*v] && 2*v <= MAX        {            dist[2*v] = dist[v] + 1;            Q.push( 2*v );        }        if( v+1 <= MAX && !dist[v+1] )        {            dist[v+1] = dist[v] + 1;            Q.push( v+1 );        }        if( !dist[v-1] && v-1 >= 0 )        {            dist[v-1] = dist[v] + 1;            Q.push( v-1 );        }    }    return dist[k]-1;}int main(){    int n, k;    while(cin >> n >> k)    cout << BFS( n, k ) << endl;    return 0;}
       
      
     
    

 

若先判断  !dist[2*v] ，可能立刻造成数组越界；若先判断  2*v <= MAX 不成立的话可以将  !dist[2*v]  短路，不会越界