bfs入门练习Catch That Cow

对于bfs宽度搜索基础练习都是大同小异，也就是说，基本上都需要一个标记数组，加上一个队列，其本质就是不停的入队，出队，直到找到结果为止，BFS对于求最短路径等问题，具有很好地优势，但是同时要注意剪枝，也就是用标记数组跳过已经检索的部分。

题目：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.

题目大意为，在一条直线上给你两个数字，有3种移动数字的方式吗，一种是坐标+1，一种是-1，一种是*2，最后要求你输出，从第一个坐标到第二个坐标所花的最短时间，每次移动算一分钟。

分析：

这显然是一个广度搜索的问题，对于每一种情况所能到达的位置入队，然后不断更新，直到找到为止。

AC代码

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
int book[1000000];
int n,k;
struct bu
{
	int x;
	int s;
};
int check(int x)
{
	if(x<0||x>1000000||book[x]==1)
	{
		return 0;
	}
	else return 1;
}
int bfs(int x)
{
	queue<bu> pi;
	bu p,q,xa;
	q.x=x;
	q.s=0;
	book[x]=1;
	pi.push(q);
	while(!pi.empty())
	{
		p=pi.front();
		pi.pop();
		if(p.x==k)
		{
			return p.s;
		}
		xa.x=p.x+1;
		if(check(xa.x))
		{
			book[xa.x]=1;
			xa.s=p.s+1;
			pi.push(xa);
		}
		xa.x=p.x-1;
		if(check(xa.x))
		{
			book[xa.x]=1;
			xa.s=p.s+1;
			pi.push(xa);
		}
		xa.x=2*p.x;
		if(check(xa.x))
		{
			book[xa.x]=1;
			xa.s=p.s+1;
			pi.push(xa);
			
		}
		
	}
	return -1;
}
int main()
{
	int sum;
	scanf("%d%d",&n,&k);
	sum=bfs(n);
	printf("%d",sum);
}