Catch That Cow (BFS)

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a pointN (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 orX + 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 andK

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：

在数轴上从n移动到k，有3种移动的方式，设当前位置的坐标为x，则一步可以到达x-1，x+1，2x的位置，

问最少经过几步可以到达指定位置

代码：

#include<stdio.h>#include<string.h>#include<queue>using namespace std;const int MAX = 100005*2;struct node{int x;int step;};int vis[MAX];int n,k;void Bfs(){struct node tem,temp;queue<node>p;temp.x=n;temp.step=0;vis[n]=1;p.push(temp);while(!p.empty()){tem=p.front();p.pop();if(tem.x==k){printf("%d\n",tem.step);return;}int x=tem.x,step=tem.step;//枚举3种方式 if(x-1>=0&&!vis[x-1]){ //左移动1位 vis[x-1]=1;temp.x=x-1;temp.step=step+1;p.push(temp);}if(x<k&&!vis[x+1]){ //右移动一位 vis[x+1]=1;temp.x=x+1;temp.step=step+1;p.push(temp);}if(x<k&&!vis[x*2]){ //向右2倍的位置 vis[x*2]=1;temp.x=x*2;temp.step=step+1;p.push(temp);}}}int main(){scanf("%d%d",&n,&k);Bfs();return 0;}