Catch That Cow（广度搜索BFS，队列）

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64

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

题意：找回丢失的牛，输入人和牛在的位置，人和牛处于一条直线上，牛不动，人可以往前往后移动一步或者一次到达它的二倍处，比如初始位置在n，下一步可以为n-1；n+1；n*2；

问走几步找到牛。

#include <iostream> 
#include <queue> 
#define SIZE 100001 
  
using namespace std; 
  
queue<int> x; 
bool visited[SIZE]; //true and false
int step[SIZE]; 
  
int bfs(int n, int k)//广搜 
{ 
    int head, next; 
    
    x.push(n); //起始节点入队 
     
    visited[n] = true; //标记n已访问 
     
    step[n] = 0; //起始步数为0 
     
    while (!x.empty()) //队列非空时 
    { 
          
        head = x.front(); //取出队头
          
        x.pop(); //弹出队头
          
        for (int i = 0; i < 3; i++) //3个方向搜索
        { 
            if (i == 0) next = head - 1; 
            else if (i == 1) next = head + 1; 
            else next = head * 2; 
              
            if (next > SIZE || next < 0) continue;//越界就不考虑了 
           
            if (!visited[next])  //判重  
            { 
                 
                x.push(next); //节点入队 
               
                step[next] = step[head] + 1; //步数+1   
                 
                visited[next] = true; //标记节点已访问 
            } 
            
            if (next == k) return step[next]; //找到退出  
        } 
    } 
} 
  
int main() 
{ 
    int n, k; 
    cin >> n >> k; 
    if (n >= k) 
    { 
        cout << n - k << endl; 
    } 
    else 
    { 
        cout << bfs(n, k) << endl; 
    } 
    return 0; 
}