【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.

 

[code][code] #include <stdio.h>

#include <iostream>

#include <vector>

#include <queue>

using namespace std ;

typedef vector<int> Table ;

const int INF = 0x7fffffff ;

const int MaxRange = 100000 ;

void run3278()

{

int i ;

int n,k ;

int range ;

int adjArr[3] ;

int vertex,adV ;

int curDist ;

queue<int> Q ;

scanf("%d%d", &n,&k) ;

//这里容易出错：Table开得太小。

//题意中已经给出最大的节点标号，所以按此数初始化Table

Table T( MaxRange+1,INF ) ;  //初始化table有MaxRange个无穷值

//BFS

T
= 0 ;  //起始点距离为0

Q.push(n) ;

while( !Q.empty() )

    {

    vertex = Q.front() ;

    Q.pop() ;

    curDist = T[vertex] ;

    //得到结果

    if( vertex == k )

   {

   printf( "%d\n" , T[vertex] ) ;

   return ;

   }

    //***得到节点v的邻接点，可能是v-1、v+1、2*v

    //超出范围的设为-1

    if( vertex-1<0 )

   adjArr[0] = -1 ;

    else

   adjArr[0] = vertex-1 ;

    if( vertex+1>MaxRange )

   adjArr[1] = -1 ;

    else

   adjArr[1] = vertex+1 ;

    if( vertex*2>MaxRange )

   adjArr[2] = -1 ;

    else

   adjArr[2] = vertex*2 ;

    //***

    //对未经处理的邻接点进行处理

    for( i=0 ; i<3 ; ++i )

   {

   if( ( adV=adjArr[i] ) == -1 )

  continue ;

   if( T[adV] == INF )

  {

  T[adV] = curDist+1 ;

  //得到结果

  if( adV == k )

 {

 printf( "%d\n" , T[adV] ) ;

 return ;

 }

  Q.push(adV) ;

  }

   }

}

}

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