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

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;}
}

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