hdu 2717 (一维广搜）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14027    Accepted Submission(s): 4293

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

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<iostream>
#include<queue>
#include<memory.h>
using namespace std;
int n,m;
int vis[100005]={0};
int dir[2]={-1,1};
struct node
{
	int x;
	int step;
}s,e;
void bfs()
{
	queue<node>q;
	node x1,t;
	s.step=0;
	vis[s.x]=1;
	q.push(s);
	while(!q.empty())
	{
		t=q.front();
		//cout<<t.x<<endl;
		q.pop();
		if(t.x==e.x)
		{
			cout<<t.step<<endl;
			return;
		}
		else
		{
			for(int i=0;i<2;i++)
			{
				x1.x=t.x+dir[i];
				if(x1.x>=0&&x1.x<=100000&&vis[x1.x]==0)
				{
					x1.step=t.step+1;
					vis[x1.x]=1;
					q.push(x1);
				}
			}
			x1.x=t.x*2;
			if(x1.x>=0&&x1.x<=100000)
			{
					x1.step=t.step+1;
					vis[x1.x]=1;
					q.push(x1);
			}
		}
		
	}
	cout<<-1<<endl;
}
int main()
{
    while(cin>>n>>m)
    { 
    	memset(vis,0,sizeof(vis));
    	s.x=n;
    	e.x=m;
    	bfs();
	}
}