Catch That Cow
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.
题意：
农夫在n点，牛在k点，找到从n到k的最短时间。
bfs代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int n,k;
struct node{int x; //位置int step; //时间
};
int vis[200000]; //标记走过的点
queue<struct node> q;
void bfs()
{while(!q.empty())q.pop(); //队列置空memset(vis,0,sizeof(vis));vis[n]=1;struct node nod={n,0}; q.push(nod);while(!q.empty()){struct node newn=q.front();q.pop();int tx;if(newn.x==k) //如果现在的位置=牛所在的位置就停止循环{printf("%d\n",newn.step);break;}for(int i=0;i<3;i++){switch(i){case 0:tx=newn.x+1;break;case 1:tx=newn.x-1;break;case 2:tx=newn.x*2;break;}if(tx>=0&&tx<=200000&&vis[tx]==0) //所在的位置符合条件就入队{struct node d={tx,newn.step+1};q.push(d);vis[tx]=1;}}}return;
}
int main()
{while(scanf("%d%d",&n,&k)!=EOF){ bfs();} return 0;
}