【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: }