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

 

 

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　　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：

 1 #include <stdio.h>

 2 #include <string.h>

 3 #include <queue>

 4 using namespace std;  5 bool isw[100010];  6 int step[100010];  7 void bfs(int n,int k)  8 {  9     memset(isw,0,sizeof(isw)); 10     queue <int> q; 11     int cur,next; 12     cur = n; 13     step[n] = 0; 14     isw[cur] = true; 15  q.push(cur); 16     while(!q.empty()){ 17         cur = q.front(); 18  q.pop(); 19         if(cur==k)    //找到，返回结果

20             return ; 21         int i; 22         for(i=1;i<=3;i++){    //步行，或者传送

23             switch(i){ 24                 case 1: 25                     next = cur - 1; 26                     if(isw[next])    //剪枝，走过的不能走

27                         break; 28                     if(next<0 || next>100010)    //剪枝，越界不能再走

29                         break; 30                     step[next] = step[cur] + 1; 31  q.push(next); 32                     isw[next] = true; 33                     break; 34                 case 2: 35                     next = cur + 1; 36                     if(isw[next]) 37                         break; 38                     if(next<0 || next>100010) 39                         break; 40                     step[next] = step[cur] + 1; 41  q.push(next); 42                     isw[next] = true; 43                     break; 44                 case 3: 45                     next = cur * 2; 46                     if(isw[next]) 47                         break; 48                     if(next<0 || next>100010) 49                         break; 50                     step[next] = step[cur] +1; 51  q.push(next); 52                     isw[next] = true; 53                     break; 54  } 55  } 56  } 57 } 58 int main() 59 { 60     int n,k; 61     while(scanf("%d%d",&n,&k)!=EOF){ 62  bfs(n,k); 63         printf("%d\n",step[k]); 64  } 65     return 0; 66 }

 

Freecode : www.cnblogs.com/yym2013