Catch That Cow（广搜）

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.
  题意是：
  给定一个起点，有三种方法能到达终点，分别是：
  1.每次减 1
  2.每次加 1

3. 每次以乘 2 前进
   求最少的步数（用BFS）
   分为两种情况：
   一个是 n>=k 就只能以第一种的方法向后退，最少步数就是两数的 差
   另一种是 n<=k 的情况
   满足范围即入队列，对当前的坐标进行标记，步数加1；到达目标点返回结果
   代码如下：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#define memset(a,n) memset(a,n,sizeof(a))
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXX = 1e5+10;

struct NODE
{
    int x;
    int step;
}now,pre,t;
int vis[MAXX];
queue<NODE> q;

int BFS(int x,int y)
{
    while(!q.empty())
        q.pop();
    pre.x = x;
    pre.step = 0;
    q.push(pre);
    vis[x] = 1;
    while(!q.empty()){
        now = q.front();
        q.pop();
        for(int i=0; i<=2; i++){
            if(i == 0)
                t.x = now.x - 1;
            else if(i == 1)
                t.x = now.x + 1;
            else if(i == 2)
                t.x = now.x * 2;
            if(t.x >= 0 && t.x <= 100000 && vis[t.x] == 0){
                if(t.x == y)
                    return now.step + 1;
                else {
                    vis[t.x] = 1;
                    t.step = now.step + 1;
                    q.push(t);
                }
            }
        }
    }

}
int main()
{
    int n,k,ans;
    memset(vis,0);
    cin >> n >> k;
    if(k <= n)
        cout << n-k << endl;
    else {
        ans = BFS(n,k);
        cout << ans << endl;
    }
}