【题解】[Usaco2007 Open]Catch That Cow 抓住那只牛-C++

题目
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?

农夫约翰被通知，他的一只奶牛逃逸了！所以他决定，马上幽发，尽快把那只奶牛抓回来,他们都站在数轴上．约翰在N(O≤N≤100000)处，奶牛在K(O≤K≤100000)处．约翰有两种办法移动，步行和瞬移：步行每秒种可以让约翰从x处走到x+1或x-1处；而瞬移则可让他在1秒内从x处消失，在2x处出现．然而那只逃逸的奶牛，悲剧地没有发现自己的处境多么糟糕，正站在那儿一动不动．那么，约翰需要多少时间抓住那只牛呢？
Input

Line 1: Two space-separated integers: N and K
仅有两个整数N和K.
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
最短的时间．
Sample Input
5 17
Farmer John starts at point 5 and the fugitive cow is at point 17.
Sample Output
4
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.
思路
这道题依然是一个BFS，可以拓展三种状态：-1，+1或*2,但是要添加一些判断使得不会让标记数组越界导致RE（我死了很多次！！！）
其他的就是模板了…

代码

 #include<bits/stdc++.h>
 using namespace std;
 int n,k;
 bool vis[];
 struct node
 {
     int x,s;
 };
 void bfs()
 {
     queue<node> q;
     vis[n]=;
     int tx;
     q.push((node){n,});
     while(!q.empty())
     {
         node now=q.front();
         //q.pop();
         if(now.x==k)
         {
             cout<<now.s<<endl;
             exit();
         }
         tx=now.x-;
         if(tx>=)
         if(!vis[tx])
         {
             q.push((node){tx,now.s+});
         }
         if(tx>*k)continue;
         tx=now.x+;
         if(!vis[tx])
         {
             q.push((node){tx,now.s+});
             vis[tx]=;
         }
         tx=now.x*;
         if(!vis[tx])
         {
             q.push((node){tx,now.s+});
             vis[tx]=;
         }
     }
 }
 int main()
 {
     cin>>n>>k;
     bfs();
     cout<<"这组数据无解"<<endl;//打着玩的
     return ;
 }