G-catch that cow

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.

题目分析

一个人站在n点，只能向左走一步或者向右走一步，或者直接走n的二倍步，要求走最小的步数抓住站在k点的牛。

解题思路

一看就是广搜的经典题目。一般要求最少步数用广搜。

源代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int  v[200005];
int cnt;
void bfs(int n,int k)
{
  queue<int>q;
  q.push(n);
  q.push(cnt);
  v[n]=1;
  while(!q.empty())
  {
    n=q.front();
    q.pop();
    cnt=q.front();
    q.pop();
    if(n==k)
    {
        return ;
    }
    if(n-1>=0&&!v[n-1])
    {
        q.push(n-1);
        q.push(cnt+1);
        v[n-1]=1;
    }
    if(n+1<200005&&!v[n+1])
    {
        q.push(n+1);
        q.push(cnt+1);
        v[n+1]=1;
    }
    if(2*n<200005&&!v[2*n])
    {
        q.push(2*n);
        q.push(cnt+1);
        v[2*n]=1;
    }
  }
}
int main()
{
  int n,k;
  while(scanf("%d%d",&n,&k)!=EOF)
  {
    if(n>=k)
    {
        printf("%d\n",n-k);
        continue;
    }
    cnt=0;
    memset(v,0,sizeof(v));
    bfs(n,k);
    printf("%d\n",cnt);
  }
  return 0;
}