传递闭包+二进制位运算+floyd（poj2570）

 
   

  
    
 
     
 
       
      

        Fiber Network 
      
 
      
 
       
 
        
 
         
 
          
Time Limit: 1000MS
 
          
 
 
          
Memory Limit: 65536K
 
         
 
         
 
          
Total Submissions: 3125
 
          
 
 
          
Accepted: 1436
 
         
 
        
 
       
 
      
 
     
 
     
Description
 
     

       Several startup companies have decided to build a better Internet, called the "FiberNet". They have already installed many nodes that act as routers all around the world. Unfortunately, they started to quarrel about the connecting lines, and ended up with every company laying its own set of cables between some of the nodes.  Now, service providers, who want to send data from node A to node B are curious, which company is able to provide the necessary connections. Help the providers by answering their queries. 
     
 
     
Input
 
     

       The input contains several test cases. Each test case starts with the number of nodes of the network n. Input is terminated by n=0. Otherwise, 1<=n<=200. Nodes have the numbers 1, ..., n. Then follows a list of connections. Every connection starts with two numbers A, B. The list of connections is terminated by A=B=0. Otherwise, 1<=A,B<=n, and they denote the start and the endpoint of the unidirectional connection, respectively. For every connection, the two nodes are followed by the companies that have a connection from node A to node B. A company is identified by a lower-case letter. The set of companies having a connection is just a word composed of lower-case letters.  After the list of connections, each test case is completed by a list of queries. Each query consists of two numbers A, B. The list (and with it the test case) is terminated by A=B=0. Otherwise, 1<=A,B<=n, and they denote the start and the endpoint of the query. You may assume that no connection and no query contains identical start and end nodes. 
     
 
     
Output
 
     

       For each query in every test case generate a line containing the identifiers of all the companies, that can route data packages on their own connections from the start node to the end node of the query. If there are no companies, output "-" instead. Output a blank line after each test case. 
       
     
 
     
Sample Input
 
     
3
1 2 abc
2 3 ad
1 3 b
3 1 de
0 0
1 3
2 1
3 2
0 0
2
1 2 z
0 0
1 2
2 1
0 0
0
 
     
Sample Output
 
     
ab
d
-

z
-
 
     题意：给出n个点，然后给出边a，b ，str代表str中的这些字母可以保证a到b的联通，然后有多组询问u和v，问u到v可以由那些字母保证连通性，若没有输出- 
    
  
    
 
     分析：一共26个字母可以使用位运算，g[a][b]代表那些可以联通，用传递闭包即可 
    
  
    
 
      
     
#include"stdio.h"
#include"string.h"
#include"stdlib.h"
#include"queue"
#include"algorithm"
#include"string.h"
#include"string"
#include"math.h"
#include"vector"
#include"stack"
#include"map"
#define eps 1e-8
#define inf 0x3f3f3f3f
#define M 250
using namespace std;
int g[M][M],vis[M];
char str[M];
int main()
{
    int n,i,a,b,c,j,k,kk=0;
    while(scanf("%d",&n),n)
    {
        memset(g,0,sizeof(g));
        memset(vis,0,sizeof(vis));
        while(scanf("%d%d",&a,&b),a||b)
        {
            scanf("%s",str);
            for(i=0;str[i]!='\0';i++)
            {
                c=str[i]-'a';
                g[a][b]|=(1<<c);
            }
        }
        for(k=1;k<=n;k++)
        {
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    g[i][j]|=(g[i][k]&g[k][j]);
                }
            }
        }
        if(kk)
            printf("\n");
        kk++;
        while(scanf("%d%d",&a,&b),a||b)
        {
            int flag=0;
            for(i=0;i<26;i++)
            {
                if(g[a][b]&(1<<i))
                {
                    printf("%c",i+'a');
                    flag++;
                }
            }
            if(flag)
            printf("\n");
            else
                printf("-\n");
        }
    }
    return 0;
}