hdu 2717:Catch That Cow（bfs广搜，经典题，一维数组搜索）

Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6383    Accepted Submission(s): 2034

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

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

USACO 2007 Open Silver

 

Recommend

teddy   |   We have carefully selected several similar problems for you:  2102 1372 1240 1072 1728 

---

 

　　bfs搜索题，基础题。

　　很简单的一道bfs搜索题，只不过把常见的二维地图换成了一维的，由于是生题，一开始想复杂了，注意剪枝，普通的广搜思路就能过。

　　代码：

 #include <stdio.h>
 #include <string.h>
 #include <queue>
 using namespace std;
 bool isw[];
 int step[];
 void bfs(int n,int k)
 {
     memset(isw,,sizeof(isw));
     queue <int> q;
     int cur,next;
     cur = n;
     step[n] = ;
     isw[cur] = true;
     q.push(cur);
     while(!q.empty()){
         cur = q.front();
         q.pop();
         if(cur==k)    //找到，返回结果
             return ;
         int i;
         for(i=;i<=;i++){    //步行，或者传送
             switch(i){
                 case :
                     next = cur - ;
                     if(isw[next])    //剪枝，走过的不能走
                         break;
                     if(next< || next>)    //剪枝，越界不能再走
                         break;
                     step[next] = step[cur] + ;
                     q.push(next);
                     isw[next] = true;
                     break;
                 case :
                     next = cur + ;
                     if(isw[next])
                         break;
                     if(next< || next>)
                         break;
                     step[next] = step[cur] + ;
                     q.push(next);
                     isw[next] = true;
                     break;
                 case :
                     next = cur * ;
                     if(isw[next])
                         break;
                     if(next< || next>)
                         break;
                     step[next] = step[cur] +;
                     q.push(next);
                     isw[next] = true;
                     break;
             }
         }
     }
 }
 int main()
 {
     int n,k;
     while(scanf("%d%d",&n,&k)!=EOF){
         bfs(n,k);
         printf("%d\n",step[k]);
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013