Catch That Cow

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27816		Accepted: 8556

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

解题报告：

这道题其实挺简单的，但是好久没有写搜索题了，很不细心，

#include 
      
        #include 
       
         #include 
        
          using namespace std; const int N = 1000010; int visit[2 *N], queue[2 * N], step[2 * N], ans;//queue起到队列的作用 void BFS(int begin, int end) {
    int rear, front, a, b[3], i;     rear = 0;//队列的尾     front = 0;//队列的头     rear ++;     queue[rear] = begin;//初始化     while (front < rear)     {
            front ++;         a = queue[front]; if (a == end)//找到         {
                ans = step[a]; return ;         } else         {
                b[0] = a + 1;             b[1] = a - 1;             b[2] = 2 * a; for (i = 0; i <=2; ++i)             {
    if (b[i] >= 0 && b[i] <= 1000000 && !visit[b[i]])//当时写的时候1000000写成了100000少写了一个零，Runtime Error好几次！！                 {
                        visit[b[i]] = 1;//标记                     rear ++;                     queue[rear] = b[i];//插到队尾                     step[b[i]] = step[a] + 1;//步数加1                 }             }         }     } } int main() {
    int  begin, end; while (scanf("%d%d", &begin, &end) != EOF)     {
            memset(visit, 0, sizeof(visit));         memset(queue, 0, sizeof(queue));         memset(step, 0, sizeof(step));         visit[begin] = 1;         BFS(begin, end);         printf("%d\n", ans);     } return 0; }
        
       
      

，哎！这道题的题意：就是在一条数轴上，求起始点到终点的最小步数，很显然得用BFS借助于队列；

代码如下：