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.
题意是：
给定一个起点，有三种方法能到达终点，分别是：
1.每次减 1 2.每次加 1 3. 每次以乘 2 前进
求最少的步数（用广搜 队列的方法）
分为两种情况：
一个是 n>=k 就只能以第一种的方法向后退，最少步数就是两数的 差
另一种是 n>=k 的情况
定义一个数组用于标记经过的位置坐标是否被访问过 就是 代码中的 flag 函数
满足范围即入队列，对当前的坐标进行标记，步数加1；
代码如下：
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<algorithm>
using namespace std;
int flag[100002];
queue<int>q;
int BFS(int n,int k)
{
int i,m,date,d;
q.push(n);//满足条件的入队列
flag[n]=1;
while(!q.empty())//一定要判断 队列是否是空队列
{
date=q.front();
q.pop();
for(i=0;i<=2;i++)
{
if(i==0)
d=date-1;//对应的三种情况
if(i==1)
d=date+1;
if(i==2)
d=date*2;
if(d>=0&&d<100001&&flag[d]==0){//flag[d]==0 代表的是没有被访问过
q.push(d);//入队列
flag[d]=flag[date]+1;// 进行到该点的时候 是第几步
if(d==k)
return flag[d]-1;/一开始定义的是 1，所以要减去 1
}
}
}
}
int main()
{
int n,k;
while(~scanf("%d %d",&n,&k))
{
memset(flag,0,sizeof(flag));
if(n>=k)
printf("%d\n",n-k);
else
printf("%d\n",BFS(n,k));//去调用 广搜函数
}
}