【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.

#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[n] = 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) ;
            }
        }
    }
}