题解报告：hdu 2717 Catch That Cow（bfs）

Problem 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

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.

解题思路：典型的bfs求最少时间。记录每个位置是否被访问，并且用一个结构体来标记每个从初始位置到达某个正访问的位置（之前为未访问）所花费的时间，有三次操作，每次操作都必须在原来的位置上进行位置转移，详解看代码。

AC代码：

 #include<iostream>

 #include<queue>

 #include<string.h>

 using namespace std;

 const int maxn=1e5+;

 int n,k,cnt;bool vis[maxn];//标记是否访问

 struct node{int x,step;}nod;//标记达到当前位置的步数

 queue<node> que;

 void bfs(int x){

     while(!que.empty())que.pop();//清空

     memset(vis,false,sizeof(vis));

     nod.x=n,nod.step=;vis[x]=true;//同时将初始位置x标记为true

     que.push(nod);//先将初始位置入队

     while(!que.empty()){

         nod=que.front();que.pop();

         if(nod.x==k){cout<<nod.step<<endl;return;}//这一步不能忘记，不然会出错，如果当前节点的值和k相等，则直接返回所花费的时间为0

         for(int i=;i<;++i){//遍历三次操作，查看是否还有可以到达的地方

             node next=nod;//每一次操作都从原来那个位置到另一个位置

             if(i==)next.x-=;

             else if(i==)next.x+=;

             else next.x*=;

             next.step++;//到达对应位置的时间在原来的基础上加1

             if(next.x==k){cout<<next.step<<endl;return;}//如果到达终点，则直接返回所花费的时间

             if(next.x>=&&next.x<maxn&&!vis[next.x]){//如果下一个位置在0~10^5范围内，并且还未访问，就可以将其入队

                 vis[next.x]=true;//将其标记为已访问状态

                 que.push(next);//将下一个位置入队

             }

         }

     }

 }

 int main(){

     while(cin>>n>>k){bfs(n);}

     return ;

 }

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