hdoj-【2717 Catch That Cow】

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12921    Accepted Submission(s): 3989

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.
需要注意有数组开大一点 
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
#define maxn 1000000
int n,k,dx[3],vis[maxn]; 
struct node
{
	int tx,time;
};
void bfs(int sx)
{
	node s;
	s.tx=sx;s.time=0;
	queue<node>que;
	que.push(s);
	vis[sx]=1;
	while(!que.empty())
	{
		node now;
		now=que.front();
		que.pop();
		if(now.tx==k)
		{
			printf("%d\n",now.time);
			return ; 
		} 
		int i,xx,temp=now.tx,tt=now.time;
		for(i=0;i<3;++i)
		{
			if(i==0)
			{
				xx=now.tx-1;
				if(xx>=0&&xx<maxn&&!vis[xx])
				{
					vis[xx]=1;now.tx=xx;now.time++;
					que.push(now); 
				} 
			}
			if(i==1)
			{
				xx=now.tx+1;
				if(xx>=0&&xx<maxn&&!vis[xx])
				{
					now.tx=xx;vis[xx]=1;now.time++;
					que.push(now); 
				} 
			} 
			if(i==2)
			{
				xx=now.tx*2;
				if(xx>=0&&xx<maxn&&!vis[xx])
				{
					now.tx=xx;now.time++;vis[xx]=1;
					que.push(now); 
				} 
			}
			now.tx=temp;now.time=tt; 
		} 
	} 
} 
int main()
{
	while(~scanf("%d%d",&n,&k))
	{
		memset(vis,0,sizeof(vis));
		bfs(n); 
	} 
	return 0;
}