POJ-3278 Catch That Cow（广搜）

POJ-3278 Catch That Cow（广搜）

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.       

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
#define MAX 100000
int k;
int book[MAX+100];
struct node
{
	int x;
	int step;
};
int cccc(int x)  
{  
    if(x>=0&&x<=MAX&&book[1]==0)  
    return 1;  
    return 0;    
    
    
    if(x<0||x>MAX||book[1])  
    return 0;  
    return 1;     //为什么或才是对的？？？？？ 
}  
int bfs(int n)
{
  queue<node> si;
  node now,next;
  now.x=n;
  now.step=0;
  book[n]=1;
  si.push(now);
  while(!si.empty())
  {
  	now=si.front();
  	si.pop();
  	if(now.x==k)
  	return now.step;
  	next.x=now.x+1;
  	if(cccc(next.x))
  	{  
  		next.step=now.step+1;
  		book[next.x]=1;
  		si.push(next);
  	}
  	next.x=now.x*2;
  		if(cccc(next.x))
  	{  
  		next.step=now.step+1;
  		book[next.x]=1;
  		si.push(next);
  	}
  	next.x=now.x-1;
  		if(cccc(next.x))
  	{  
  		next.step=now.step+1;
  		book[next.x]=1;
  		si.push(next);
  	}
  } 
}
int main()
{
	int n;
	int res;
	memset(book,0,sizeof(book));
    scanf("%d%d",&n,&k);
	res=bfs(n);
	printf("%d\n",res);
	return 0;
}