[题解] Catch That Cow 抓住那只牛 C++

题目

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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来,他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有两种办法移动，步行和瞬移：步行每秒种可以让约翰从x处走到x+1或x-1处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．那么，约翰需要多少时间抓住那只牛呢？

Input

• Line 1: Two space-separated integers: N and K
  仅有两个整数N和K.

Output

• Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
  最短的时间．

Sample Input

5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.

Sample Output

4
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.

思路

典型bfs板子题,且是在一维数组上进行，每次for0~1扩展前进1布和后退1布，再扩展2x步，判断是否出界标记
要是数组刚好100000，就算你判断了边界也会RE,再多开一半就没事了

代码

#include
using namespace std;
int n,k;
int dir[2]={1,-1};
struct node{
	int x;
	int d;
	node(int xx,int dd){
		x=xx;
		d=dd;
	}
};
bool in(int x){
	return x>=0 && x<=150000;
}
bool vis[150005];
int bfs(int sx){
	queue<node> q;
	if(sx==k){
		return 0;
	}
	q.push(node(sx,0));
	while(!q.empty()){
		node now=q.front();
		//q.pop();
		for(int i=0;i<2;i++){
			int tx=now.x+dir[i];
			if(in(tx) && !vis[tx]){
				vis[tx]=1;
				if(tx==k){
					return now.d+1;
				}
				q.push(node(tx,now.d+1)); 
			}	
		}
		int tx=2*now.x;
		if(in(tx) && !vis[tx]){
			vis[tx]=1;
			if(tx==k){
				return now.d+1;
			}
			q.push(node(tx,now.d+1));
		} 
	}
}
int main(){
	while(1);
	string s;
	cin>>n>>k;
	getline(cin,s);
	cout<<bfs(n);
	return 0;
}