B - Catch That Cow（广度搜索）

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我是靠谱客的博主畅快钢笔，最近开发中收集的这篇文章主要介绍B - Catch That Cow（广度搜索），觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (0 ≤N ≤ 100,000) on a number line and the cow is at a pointK (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 orX+ 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

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.

本题思路：

哎我是copy的

主要在于利用队列这个广搜

c++代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=100009;
int book[maxn]={0};
int step[maxn]={0};
queue<int>q;
int bfs(int n,int k)
{
   int i;
   int t;
   book[n]=1;
   step[n]=0;
   q.push(n);
   while(q.empty()==0)
   {
       int head=q.front();
       q.pop();
       for(i=1;i<=3;i++)
       {
           if(i==1)
           t=head+1;
           else if(i==2)
           t=head-1;
           else if(i==3)
           t=head*2;
           if(t<0||t>maxn)
           continue;
           if(book[t]==0)
           {
           q.push(t);
           book[t]=1;
           step[t]=step[head]+1;
           }
           if(t==k)
           return step[t];
        }
   }
}
int main()
{
   int n,k;
   while(~scanf("%d%d",&n,&k))
   {
       memset(book,0,sizeof(book));
          memset(step,0,sizeof(step));
       while(q.empty()==0)q.pop();
       if(n>=k)printf("%dn",n-k);
       else if(n<k)printf("%dn",bfs(n,k));
   }
   return 0;
}