【原】 POJ 3278 Catch That Cow BFS单源无权图最短距离 解题报告

 

http://poj.org/problem?id=3278

 

方法:
单源无权图最短距离,即BFS
该问题只需要求得某两点间的最短距离,所以不必求得所有节点的最短距离,一旦处理了目的地点,即可返回结果

 

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.

 

   1: #include <stdio.h>
   2: #include <iostream>
   3: #include <vector>
   4: #include <queue>
   5:  
   6: using namespace std ;
   7:  
   8: typedef vector<int> Table ;
   9: const int INF = 0x7fffffff ;
  10: const int MaxRange = 100000 ;
  11:  
  12: void run3278()
  13: {
  14:     int i ;
  15:     int n,k ;
  16:     int range ;
  17:     int adjArr[3] ;
  18:     int vertex,adV ;
  19:     int curDist ;
  20:     queue<int> Q ;
  21:  
  22:     scanf("%d%d", &n,&k) ;
  23:  
  24:     //这里容易出错:Table开得太小。
  25:     //题意中已经给出最大的节点标号,所以按此数初始化Table
  26:     Table T( MaxRange+1,INF ) ;  //初始化table有MaxRange个无穷值
  27:     
  28:     //BFS
  29:     T[n] = 0 ;  //起始点距离为0
  30:     Q.push(n) ;
  31:     while( !Q.empty() )
  32:     {
  33:         vertex = Q.front() ;
  34:         Q.pop() ;
  35:         curDist = T[vertex] ;
  36:  
  37:         //得到结果
  38:         if( vertex == k )
  39:         {
  40:             printf( "%d\n" , T[vertex] ) ;
  41:             return ;
  42:         }
  43:  
  44:         //***得到节点v的邻接点,可能是v-1、v+1、2*v
  45:         //超出范围的设为-1
  46:         if( vertex-1<0 )
  47:             adjArr[0] = -1 ;
  48:         else
  49:             adjArr[0] = vertex-1 ;
  50:  
  51:         if( vertex+1>MaxRange )
  52:             adjArr[1] = -1 ;
  53:         else
  54:             adjArr[1] = vertex+1 ;
  55:  
  56:         if( vertex*2>MaxRange )
  57:             adjArr[2] = -1 ;
  58:         else
  59:             adjArr[2] = vertex*2 ;
  60:         //***
  61:  
  62:         //对未经处理的邻接点进行处理
  63:         for( i=0 ; i<3 ; ++i )
  64:         {
  65:             if( ( adV=adjArr[i] ) == -1 )
  66:                 continue ;
  67:             if( T[adV] == INF )
  68:             {
  69:                 T[adV] = curDist+1 ;
  70:  
  71:                 //得到结果
  72:                 if( adV == k )
  73:                 {
  74:                     printf( "%d\n" , T[adV] ) ;
  75:                     return ;
  76:                 }
  77:  
  78:                 Q.push(adV) ;
  79:             }
  80:         }
  81:     }
  82: }
posted @ 2010-11-08 19:46  Allen Sun  阅读(317)  评论(0编辑  收藏  举报