Catch That Cow POJ - 3278

Catch That Cow POJ - 3278

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.

刚写的时候数组越界，debug发现没有在没有考虑位置为负的情况，还有要先判断是否越界在判断是否访问过。

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;

const int maxn=100010;
int n,k;
int vis[maxn];
int ans;

struct node{
	int x;
	int step;
};

bool judge(int x)
{

	if(x<0||x>=maxn)
	{
		return true;
	}
	if(vis[x])
	{
		return true;
	}
	return false;
}

void bfs()
{
	node cur,next;
	queue<node>q;
	cur.x=n;
	cur.step=0;
	q.push(cur);
	vis[n]=true;
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		int i,nx;
		if(cur.x==k)
		{
			ans=cur.step;
			return;
		}
		next.x=cur.x+1;
		if(!judge(next.x))
		{
			vis[next.x]=true;
			next.step=cur.step+1;
			q.push(next);
		}
		next.x=cur.x-1;	
		if(!judge(next.x))
		{
			vis[next.x]=true;
			next.step=cur.step+1;
			q.push(next);		
		}	
		next.x=cur.x*2;
		if(!judge(next.x))
		{
			vis[next.x]=true;
			next.step=cur.step+1;
			q.push(next);
		}
	}
}

int main()
{
	cin>>n>>k;
	memset(vis,0,sizeof(vis));
	bfs();
	cout<<ans;
	return 0;
}