poj Catch That Cow bfs

最新推荐文章于 2019-10-14 21:22:18 发布

xiaoleiacm

最新推荐文章于 2019-10-14 21:22:18 发布

阅读量401

点赞数

CC 4.0 BY-SA版权

分类专栏：广度优先搜索

本文链接：https://blog.youkuaiyun.com/xiaoleiacmer/article/details/8552809

广度优先搜索专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一个基于宽度优先搜索（BFS）的经典算法问题——如何最短时间捕捉逃跑的奶牛。该问题通过模拟农夫的不同移动方式来寻找最优路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 33812		Accepted: 10419

Description

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 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 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.

Source

USACO 2007 Open Silver

题意：给定两个数，从第一个数到第二个数最少经过几步可以到达，bfs进行搜索

这个是一维数组进行bfs 学习了一下，思路都差不多

其实bfs 和dfs就是树的问题

代码：

#include<iostream>
#include<queue>
#include<algorithm>
using namespace std;
#define maxn 100001 
queue<int> x;
bool vis[maxn];
int step[maxn];
int bfs(int n,int k)
{
	int head,next;
	x.push(n);
	vis[n]=1;
	step[n]=0;
	while(!x.empty())
	{
		head=x.front();
		x.pop();
		for(int i=0;i<3;i++)
		{
			if(i==0) 
			 next=head+1;
			else if(i==1)
			 next=head-1;
			else if(i==2)
			 next=head*2;
			
			if(next>maxn||next<0)
			continue;
			
			if(!vis[next])
			{
				x.push(next);
				step[next]=step[head]+1;
				vis[next]=1;
			}
			if(next==k) return step[next];
			
		}
		
	}
}
int main()
{
	int n,k;
	cin>>n>>k;
	if(n>=k)
	{
		cout<<n-k<<endl;		
	}
	else
	{
		cout<<bfs(n,k)<<endl;
	}
	return 0;
}