HDU2717 Catch That Cow (BFS)

Description

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.

经典BFS。对所有情况进行一次搜索，最先找到的就是解。

代码如下：

#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
const int N=1000000;
int map[N+10],n,k;

struct point
{
	int x,step;
} ;
 
int check(int x)
{
 	if(x<0||x>N||map[x])
 	{
 		return 0; 
	}
	return 1;
}
 
int bfs(int x)
{
	queue<point> q;  //队列 
	point now,next;
	now.x=x;now.step=0; 
	map[x]=1;  //标记 
	q.push(now); //将now插入队列
	while(!q.empty())
	{
		now=q.front(); //将先进入队列的元素取出存入now
		q.pop(); //取出后删除
		if(now.x==k) return now.step;
		next=now;
		next.x=now.x+1;
		//三种情况 
		if(check(next.x))
		{ 
			next.step=now.step+1; //步数加一； 
			map[next.x]=1;//标记 
			q.push(next);
		}
		next.x=now.x-1;
		if(check(next.x))
		{
			next.step=now.step+1;
			map[next.x]=1;
			q.push(next);
		}
		next.x=now.x*2;
		if(check(next.x))
		{
			next.step=now.step+1;
			map[next.x]=1;
			q.push(next);
		}
	 }
	 return -1; 
}
 
int main()
{
	scanf("%d%d",&n,&k);
	memset(map,0,sizeof(map));//标记为0（表示未搜索过）
	cout<<bfs(n)<<endl;
	return 0; 
}

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