POJ 2570 Fiber Network

POJ 2570 Fiber Network

题目大意是，某条道路由一些公司修建，修建道路的公司可以提供这条路上的连通，询问哪些公司可以提供从A到B的路径。
类似于Floyed的算法。用二进制数表示一个集合。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/2 15:57:29
 * File Name: poj2570.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  L(x) ((x)<<1)
#define  R(x) (((x)<<1)+1)
#define  Half(x) ((x)>>1)
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
typedef  long   long  int64;
typedef unsigned  long   long  uint64;
const   int  kMaxn( 207 );

int  n,cnt,d[kMaxn][kMaxn];
map < string , int >  id;

int  main()
{
     // freopen("data.in","r",stdin);
    
     while (scanf( " %d " , & n) == 1   &&  n)
    {
        cnt = 0 ;
        memset(d, 0 , sizeof (d));
         int  u,v;
         while (scanf( " %d%d " , & u, & v) == 2   &&  (u  ||  v))
        {
             string  company;
            cin >> company;
             for ( int  i = 0 ;i < company.size();i ++ )
                d[u][v] |= ( 1 << (company[i] - ' a ' ));
        }
         for ( int  k = 1 ;k <= n;k ++ )
             for ( int  i = 1 ;i <= n;i ++ )
                 for ( int  j = 1 ;j <= n;j ++ )
                    d[i][j] |= d[i][k] & d[k][j];
         while (scanf( " %d%d " , & u, & v) == 2   &&  (u  ||  v))
        {
             bool  found( false );
             for ( int  i = 0 ;i < 26 ;i ++ )
                 if (d[u][v] & ( 1 << i))
                {
                    printf( " %c " ,( char )(i + ' a ' ));
                    found = true ;
                }
             if ( ! found)
                printf( " - " );
            printf( " \n " );
        }
        printf( " \n " );
    }
    
     return   0 ;
}