Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9276    Accepted Submission(s):
2907

Problem 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

题意：人在n位置处，牛在k位置处，牛不动问人最短多少分钟可以捉到牛，人有两种移动方法

1、一分钟移动一步n-1或者n+1

2、一分钟移动2*n步；

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 1000100
using namespace std;
int n,m;
int vis[MAX];
struct node
{
	int x;
	int step;
	friend bool operator < (node a,node b)
	{
		return a.step>b.step;
	}
};
int judge(int x)
{
	if(x < 0||x > MAX||vis[x])
	    return 0;
	return 1;
}
void bfs(int n,int k)
{
	int i,j;
	priority_queue<node>q;
	node beg,end;
	beg.x=n;
	beg.step=0;
	q.push(beg);
	vis[n]=1;
	while(!q.empty())
	{
		beg=q.top();
		q.pop();
		if(beg.x==k)
		{
			printf("%d\n",beg.step);
			return ;
		}
		end.x=beg.x-1;
		if(judge(end.x))
		{
			vis[end.x]=1;
			end.step=beg.step+1;
			q.push(end);
		}
		end.x=beg.x+1;
		if(judge(end.x))
		{
			vis[end.x]=1;
			end.step=beg.step+1;
			q.push(end);
		}
		end.x=beg.x*2;
		if(judge(end.x))
		{
			vis[end.x]=1;
			end.step=beg.step+1;
			q.push(end);
		}
	}
}
int main()

{
	int i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		bfs(n,m);
	}
	return 0;
}