POJ 3278 Catch That Cow（模板——BFS）

题目链接：
http://poj.org/problem?id=3278

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.

题意描述：
输入人和牛在坐标轴上的位置

按照人寻找的三种走路方式，问最短抓到牛的时间。

解题思路：

搜索题，求最短时间，使用BFS更合适一点。另外需要注意：

 　　走过的点是不需要重复走的，因为既然走过就证明牛不在这个点，所以使用book数组标记一下就不会出现超内存（扩展无用点）和超时啦。

AC代码：

 #include<stdio.h>

 int bfs(int n,int k);
 struct node
 {
     int x,s;
 };
 struct node q[];
 int book[];//标记数组，否则超内存 

 int main()
 {
     int n,k;
     while(scanf("%d%d",&n,&k) != EOF)
     {
         if(n==k)
         printf("0\n");
         else
         printf("%d\n",bfs(n,k));
     }
     return ;
 }
 int bfs(int n,int k)
 {
     int i,head,tail,tx;
     head=;
     tail=;
     q[tail].x=n;
     q[tail].s=;
     book[n]=;
     tail++;

     while(head < tail)
     {
         for(i=;i<=;i++)
         {
             if(i==)
             tx=q[head].x-;
             if(i==)
             tx=q[head].x+;
             if(i==)
             tx=q[head].x*;

             if(tx <  || tx > )
                 continue;
             if(!book[tx])
             {
                 book[tx]=;
                 q[tail].x=tx;
                 q[tail].s=q[head].s+;
                 tail++;

                 if(tx == k)
                     return q[tail-].s;
             }
         }
         head++;
     }
 }