poj 2570 floyd+二进制

二进制真的是个神奇的东西！

类似floyd求传递闭包，这里g[i][j]表示i点到j点的状态（即符合条件的公司的集合）。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5
 6 const int N = 201;
 7 const int M = 31;
 8 int g[N][N];
 9 int n;
10
11 int main ()
12 {
13     while ( scanf("%d", &n), n )
14     {
15         memset( g, 0, sizeof(g) );
16         int u, v;
17         char str[M];
18         while ( scanf("%d%d", &u, &v) != EOF )
19         {
20             if ( !u && !v ) break;
21             scanf("%s", str);
22             int len = strlen(str);
23             for ( int i = 0; i < len; i++ )
24             {
25                 int tmp = str[i] - ‘a‘;
26                 g[u][v] |= ( 1 << tmp );
27             }
28         }
29         for ( int k = 1; k <= n; k++ )
30         {
31             for ( int i = 1; i <= n; i++ )
32             {
33                 for ( int j = 1; j <= n; j++ )
34                 {
35                     g[i][j] |= ( g[i][k] & g[k][j] );
36                 }
37             }
38         }
39         int p, q;
40         while ( scanf("%d%d", &p, &q) != EOF )
41         {
42             if ( !p && !q ) break;
43             if ( !g[p][q] )
44             {
45                 putchar(‘-‘);
46             }
47             else
48             {
49                 for ( int i = 0; ( 1 << i ) <= g[p][q]; i++ )
50                 {
51                     if ( g[p][q] & ( 1 << i ) )
52                     {
53                         putchar( ‘a‘ + i );
54                     }
55                 }
56             }
57             putchar(‘\n‘);
58         }
59         putchar(‘\n‘);
60     }
61     return 0;
62 }