Catch That Cow(广度优先搜索_bfs)



Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 48036		Accepted: 15057

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

题意：输入两个数n，k。求从n到k最少走多少步。能够前进1后退1或者当前的位置*2。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
struct node
{
    int x;//当前位置
    int ans;//走的步数
}q[1000010];
int vis[1000010];//标记变量，该点是否被訪问；
int jx[]={-1,1};//后退1或者前进1。
struct node t,f;
int n,k;
void bfs()
{
     int i;
     int s=0,e=0;//指针模拟队列。
往队列加e++ 往队列里提出数s++
     memset(vis,0,sizeof(vis));
     t.x=n;//当前初始位置
     vis[t.x]=1;//标记为1代表訪问过。
     t.ans=0;//初始位置步数为0；
     q[e++]=t;//把当前步数加人队列
     while(s<e)//当队列不为空
    {
        t=q[s++];//提出
        if(t.x==k)//假设该数正好等于目标位置直接输出步数
        {
             printf("%d
",t.ans);
             break;
        }
        for(i=0;i<3;i++)//i=0后退一步，i=1前进一步。i=2此时的位置*2；
        {
             if(i==2)
             {
                 f.x=t.x*2;
             }
             else
             {
                 f.x=t.x+jx[i];
             }
             if(f.x>=0&&f.x<=100000&&!vis[f.x])
             {
                 f.ans=t.ans+1;
                 q[e++]=f;
                 vis[f.x]=1;
             }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        bfs();
    }
    return 0;
}