【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[10000010];
struct node
{
    int x,s;
};
void bfs()
{
    queue<node> q;
    vis[n]=1;
    int tx;
    q.push((node){n,0});
    while(!q.empty())
    {
        node now=q.front();
        //q.pop();
        if(now.x==k)
        {
            cout<<now.s<<endl;
            exit(0);
        }
        tx=now.x-1;
        if(tx>=0)
        if(!vis[tx])
        {
            q.push((node){tx,now.s+1});
        }
        if(tx>2*k)continue;
        tx=now.x+1;
        if(!vis[tx])
        {
            q.push((node){tx,now.s+1});
            vis[tx]=1;
        }
        tx=now.x*2;
        if(!vis[tx])
        {
            q.push((node){tx,now.s+1});
            vis[tx]=1;
        }
    }
}
int main()
{
    cin>>n>>k;
    bfs();
    cout<<"这组数据无解"<<endl;//打着玩的
    return 0;
}