Catch That Cow
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 88732 | Accepted: 27795 |
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.
Source
题目链接：http://poj.org/problem?id=3278
题意：有一个农民和一头牛，他们在一个数轴上，牛在k位置保持不动，农户开始时在n位置。设农户当前在M位置，每次移动时有三种选择：1.移动到M-1；2.移动到M+1位置；3.移动到M*2的位置。问最少移动多少次可以移动到牛所在的位置。所以可以用广搜来搜索这三个状态，直到搜索到牛所在的位置。
分析：BFS的板子题，用队列做，第一次学队列，感觉自己好弱啊！现在才学，算法真的好难学，学了又怕不会用，现在不敢学太快了！
下面每一步代码我都给出了详细解释，一看就懂！
1 #include2 #include 3 #include 4 #include 5 #include 6 #define MAX 100001 7 using namespace std; 8 queue q;//使用前需定义一个queue变量，且定义时已经初始化 9 bool visit[MAX];//访问空间10 int step[MAX]; //记录步数的数组不能少11 bool bound(int num)//定义边界函数12 {13 if(num<0||num>100000)14 return true;15 return false;16 }17 int BFS(int st,int end)18 {19 queue q;//使用前需定义一个queue变量，且定义时已经初始化20 int t,temp;21 q.push(st);//进队列22 visit[st]=true;23 while(!q.empty())//重复使用时，用这个初始化24 {25 t=q.front();//得到队首的值26 q.pop();//出队列,弹出队列的第一个元素，并不会返回元素的值27 for(int i=0;i<3;++i) //三个方向搜索28 {29 if(i==0)30 temp=t+1;31 else if(i==1)32 temp=t-1;33 else34 temp=t*2;35 if(bound(temp)) //越界36 continue;37 if(!visit[temp])//访问空间未被标记38 {39 step[temp]=step[t]+1;40 if(temp==end)41 return step[temp];42 visit[temp]=true;//标记此点43 q.push(temp);//将temp元素接到队列的末端；44 }45 }46 }47 }48 int main()49 {50 int st,end;51 while(scanf("%d%d",&st,&end)!=EOF)52 {53 memset(visit,false,sizeof(visit));//visit数组进行初始化操作54 if(st>=end)55 cout< <