C - 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 ：一头牛从牛栏跑了出来，现在要把它抓回来，给你人的位置和牛的位置，人有三种移动方式，一种是当前坐标减一，一种是当前坐标加一，还有一种是当前坐标乘以二，三种移动方式耗费的时间是一样的，假设牛没有发现人要抓它，不移动，问要多长时间人抓到牛。

解题思路 ：广搜，首先将人的位置加入队列，然后分三种情况进行判断。

#include<stdio.h>
#include<string.h>

int n,k;
int book[100100];

struct node
{
	int x,s;
}que[100000],q;

int bfs()
{
	int head,tail,i;
	head = 0;
	tail = 0;
	que[tail].x = n;
	que[tail].s = 0;
	tail ++;
	book[n] = 1;
	if(n >= k)
	return n-k;
	
	while(head < tail)
	{
		q = que[head ++];
		
		if(q.x+1 > 0&&q.x+1 < 100100&&book[q.x+1] == 0)
		{
			book[q.x+1] = 1;
			que[tail].x = q.x + 1;
			que[tail].s = q.s + 1;
			tail ++;
			if(q.x+1 == k)
			return q.s + 1;
		}
		if(q.x-1 > 0&&q.x-1 < 100100&&book[q.x-1] == 0)
		{
			book[q.x-1] = 1;
			que[tail].x = q.x - 1;
			que[tail].s = q.s + 1;
			tail ++;
			if(q.x-1 == k)
			return q.s + 1;
		}
		if(q.x*2 > 0&&q.x*2 < 100100&&book[q.x*2] == 0)
		{
			book[q.x*2] = 1;
			que[tail].x = q.x*2;
			que[tail].s = q.s + 1;
			tail ++;
			if(q.x*2 == k)
			return q.s + 1;
		}
	}
}
int main()
{
	int t;
	while(~ scanf("%d %d",&n,&k))
	{
		memset(book,0,sizeof(book));
		t = bfs();
		printf("%d\n",t);
	}
	return 0;
}