C - Catch That Cow（bfs的简单应用）

Catch That Cow

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：给定农夫约翰的和母牛的位置，其中农夫可以进行两种操作：
*步行：农夫约翰可以在一分钟内从任意一个点移动到X-1或X+1
*心灵传送：农夫约翰可以在一分钟内从任意一个点移动到2个点。而母牛不动，则对于农夫约翰需要多长时间才能找回他？

解题思路：此题也是求最早寻找时间，我们则可以利用bfs搜索。此题易解。

AC代码：

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<memory.h>

using namespace std;

int n,k;                 //n代表农夫约翰的位置，k代表母牛的位置。
const int maxn=1e5+5;
bool visited[maxn];     //辅助数组，判断是否被访问过
int result[maxn];       //存储步数

void bfs(int n,int k)//通过bfs来寻求最短路径。
{
	queue<int> Q;       //利用队列
	int head,temp;
	head=n;
	visited[n]=true;
	Q.push(head);
	while(!Q.empty())
	{
		head=Q.front();
		Q.pop();
		for(int i=0;i<3;i++)    //三种步骤
		{
			temp=head;
			if(i==0)temp+=1;
			else if(i==1)temp-=1;
			else temp*=2;
			if(temp<0||temp>maxn)continue;  //判断是否出界
			if(!visited[temp])             //判断是否被访问过
			{
				result[temp]=result[head]+1;
				Q.push(temp);
				visited[temp]=true;
			}
			if(temp==k){cout<<result[temp]<<endl;return;}
		}
	}
	return ;
}
int main()
{
	while(cin>>n>>k)
	{
		if(n>=k){cout<<n-k<<endl;continue;}  //考虑n大于等于k的情况，此时n只能往后减一
		memset(visited,0,sizeof(visited));
		memset(result,0,sizeof(result));
		bfs(n,k);
	}
}