Catch That Cow

Time Limit: 2000MS      Memory Limit: 65536K

BFS

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?

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：

有一个牧羊人和一头牛，人在x点，牛在y点，他们都在一条数轴上，牧羊人想要去抓牛，但是他一次只能走到x+1，x-1或者是2*x处；在抓牛的过程中牛不动，问你牧羊人最短要多长时间能够抓到牛，人跳一次花一分钟；

解题思路：

把每种情况都走一遍，谁最先到达谁就是最短的，走过了的就标记一下；

完整代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
const int maxn=200000;////要开打两倍！！！估计是因为有2*x吧？？！！。。
int vis[200010];
int ans;
struct node
{
    int x;
    int num;
};
int bfs(int st,int en)
{
    queue<node>q;
    node now,next;
    now.x=st;
    now.num=0;
    vis[st]=1;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(now.x==en)
        {
            ans=now.num;
            break;
        }
        next=now;
        next.x=now.x+1;
        if(next.x>=0&&next.x<maxn&&vis[next.x]==0)
        {
            next.num=now.num+1;
            vis[next.x]=1;
            q.push(next);
        }
        next.x=now.x-1;
        if(next.x>=0&&next.x<maxn&&vis[next.x]==0)
        {
            next.num=now.num+1;
            vis[next.x]=1;
            q.push(next);
        }
        next.x=2*now.x;
        if(next.x>=0&&next.x<maxn&&vis[next.x]==0)
        {
            next.num=now.num+1;
            vis[next.x]=1;
            q.push(next);
        }
    }
    return ans;
}
int  main()
{
    int st,en;
    while(scanf("%d%d",&st,&en)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        printf("%d\n", bfs(st,en));
    }
    return 0;
}