bfs

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.

 

思路：

　　Bfs

代码如下：

　　

#include <iostream>
#include <queue>
#define MAX 100001
using namespace std;

int vis[MAX];
int ret[MAX];
queue <int> s;

int bfs(int a, int b)
{
 if(a == b)
 {
  return 0;
 }
 s.push(a);
 int cur;
 while (!s.empty())
 {
  cur = s.front();
  s.pop();
  if(cur + 1 < MAX && !vis[cur + 1])
  {
   s.push(cur + 1);
   ret[cur + 1] = ret[cur] + 1;
   vis[cur + 1] = 1;
  }
  if(cur + 1 == b)
  {
   break;
  }
  if(cur - 1 >= 0 && !vis[cur - 1])
  {
   s.push(cur - 1);
   ret[cur - 1] = ret[cur] + 1;
   vis[cur - 1] = 1;
  }
  if(cur - 1 == b)
  {
   break;
  }
  if(cur << 1 < MAX && !vis[cur << 1])
  {
   s.push(cur << 1);
   ret[cur << 1] = ret[cur] + 1;
   vis[cur << 1] = 1;
  }
  if(cur << 1 == b)
  {
   break;
  }
 }
 return ret[b];
}

int main()
{
 int a, b;
 cin >> a >> b;
 cout << bfs(a, b) << endl;
 return 0;
}