5.12 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 100001using 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;}

　　 

转载于:https://www.cnblogs.com/lihek/archive/2013/05/12/3074186.html