J - Catch That Cow POJ - 3278 （BFS）

题目：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.
BFS入门

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

const int maxn=100001;

bool vis[maxn];//标记数组
int step[maxn];//记录到了每一位置所走的步数
queue <int> q;//定义队列

int bfs(int n,int k) {
	int head,next;//head 表示准备走时候的位置 next表示下一时刻位置 
	q.push(n);   //开始FJ在n位置，n入队
	step[n]=0;  //记录步数  
	vis[n]=true; //标记已访问 如果标记了 就不走 避免死循环 
	while(!q.empty()) { //当队列非空
		head=q.front();  //取队首
		q.pop();         //弹出对首
		for(int i=0; i<3; i++) { //FJ的三种走法·
			if(i==0) next=head-1;//-1走法 
			else if(i==1) next=head+1;//+1走法 
			else next=head*2;//*2走法 
			if(next<0 || next>=maxn) continue; //排除出界情况
			if(!vis[next]) { //如果next位置未被访问 q数组表示当前人的位置 
				q.push(next);    //入队 更新人的位置 
				step[next]=step[head]+1;  //步数+1
				vis[next]=true;  //标记已访问
			}
			if(next==k) return step[next];  //当遍历到结果，返回步数
		}
	}
}
int main() {
	int n,k;
	while(cin>>n>>k) {
		memset(step,0,sizeof(step));
		memset(vis,false,sizeof(vis));

		while(!q.empty()) 
		q.pop(); //注意调用前要先清空
		if(n>=k) printf("%d\n",n-k);
		else printf("%d\n",bfs(n,k));
	}
	return 0;
}