HDU2717 Catch That Cow

                                                         Catch That Cow

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4945    Accepted Submission(s): 1567

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


解 题思路，直接广度优先搜索，搜索方向作变动，有传统的4个相邻方向变成一维相邻，或飞跃。注意空间的大小，别爆栈或超内存。

#include<cstdio> #include<cstring> #include<queue> using namespace std; int map[1000000]; int flood[1000000]; int dir[2]={1,-1}; int n,s,e; struct node {   int x;   int step; }; int bfs() {     node now,next;     queue<node> q;     now.x=s;     now.step=0;     flood[s]=1;     if(e==s)         return 0;     q.push(now);     while(!q.empty())     {         now=q.front();         q.pop();         for(int i=0;i<2;i++)      //往上或往下一层         {             next.x=now.x+dir[i];             next.step=now.step+1;             if(next.x>=0&&next.x<1000000&&!flood[next.x])             {                 if(next.x==e)                     return next.step;                 flood[next.x]=1;                 q.push(next);             }         }         next.x=now.x*2;      //飞跃到2*x点去         next.step=now.step+1;         if(next.x>=0&&next.x<1000000&&!flood[next.x]) //不要爆栈         {             if(next.x==e)                 return next.step;             flood[next.x]=1;             q.push(next);         }     }     return -1; } int main() {     int i;     while(scanf("%d%d",&s,&e)!=EOF)     {         memset(flood,0,sizeof(flood));         printf("%d\n",bfs());     }     return 0; }