C - Catch That Cow

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最新推荐文章于 2025-06-26 15:06:27 发布

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博客讲述农夫 John 要在数轴上抓住不动的牛，他有步行和瞬移两种移动方式，步行可每分钟从 X 到 X - 1 或 X + 1，瞬移可每分钟从 X 到 2 × X。给出农夫和牛的初始位置，求抓住牛的最少时间，并给出了示例输入输出。

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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.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
int visited[100010];
typedef struct
{
	int a;
	int stap;
}SE;
SE s,e;
queue<SE>q;
int bfs(int n,int k)
{
	s.a=n;
	s.stap=0;
    q.push(s);
	while(q.size())
	{
		s=q.front();
		q.pop();
		if(s.a==k)
		return s.stap;
		visited[s.a]=1;
		if(s.a+1>=0&&s.a+1<=100000&&!visited[s.a+1])
		{
			e.a=s.a+1;
			e.stap=s.stap+1;
			q.push(e);
			visited[s.a+1]=1;
		}
		if(s.a-1>=0&&s.a-1<=100000&&!visited[s.a-1])
		{
			e.a=s.a-1;
			e.stap=s.stap+1;
			q.push(e);
			visited[s.a-1]=1;
		}
		if(s.a*2>=0&&s.a*2<=100000&&!visited[s.a*2])
		{
			e.a=s.a*2;
			e.stap=s.stap+1;
			q.push(e);
			visited[s.a*2]=1;
		}
	}
}
int main()
{
   int n,k;
   cin>>n>>k;
   cout<<bfs(n,k)<<endl;
	return 0;
}