求强连通分量

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstdlib>
  4 #include <cstring>
  6 #include <cstdio>
  7 #include <vector>
  8 #include <queue>
  9 using namespace std ;
 10 const int inf = 1 << 30 , maxn = 10000 + 100 , maxe = 500000 + 100 ;
 11 int  n , m , cnt ,head[maxn] , tot  , top , sta[maxn], dfn[maxn] , low[maxn] , sum[maxn];
 12 struct id {
 13     int to , nxt ;
 14 }  edge[maxe] ;
 15 bool lie[maxe] ;
 16 vector< vector<int> > node( maxn ) ; 
 17 
 18 
 19 inline void Init(  ) {
 20     freopen( "10717.in" , "r" , stdin ) ;
 21     freopen( "10717.out" , "w" , stdout ) ;
 22 }
 23 
 24 int read(  ) {
 25     char ch = getchar(  ) ; int ret = 0 , k = 1 ;
 26     while( ch < ‘0‘ || ch > ‘9‘ ) { if( ch == ‘-‘ ) k = -1 ; ch = getchar(  ) ; }
 27     while( ch >= ‘0‘ && ch <= ‘9‘ ) ret = ret * 10 + ch - ‘0‘ , ch = getchar(  ) ;
 28     return ret * k ;
 29 }
 30 
 31 
 32 void add( int u , int v ) {
 33     edge[++cnt].to = v ;
 34     edge[cnt].nxt = head[u] ;
 35     head[u] = cnt ;
 36 }
 37 
 38 
 39 void input(  ) {
 40     n = read(  ) , m = read(  ) ;
 41     int  u , v ;
 42     memset( head , -1 , sizeof( head ) ) ;
 43     for( int x = 1 ;  x <= m ; ++x ) {
 44         u = read( ) , v = read(  ) ;
 45         add( u , v ) ;
 46     }
 47     cnt = 0 ;
 48 }
 49 
 50 
 51 void scc( int u ) {
 52     dfn[u] = ++tot ; low[u] = tot ;
 53     sta[++top] = u ; lie[u] = true ;
 54     for( int i = head[u] ; ~i ; i = edge[i].nxt ) {
 55         int v = edge[i].to ;
 56         if( dfn[v] == 0 ) 
 57         {
 58             scc( v ) ;
 59             low[u] = min( low[u] , low[v] ) ;
 60         }
 61         else if( lie[v] ) {
 62             low[u] = min( low[u] , dfn[v] ) ;
 63         }    
 64     }
 65     if( low[u] == dfn[u] ) {
 66         cnt ++ ;
 67         int front = 0 ;
 68         do {
 69             front = sta[top--] ;
 70             sum[cnt] ++ ;
 71             node[cnt].push_back( front ) ;
 72             lie[front] = false ;
 73             
 74         } while( front != u ) ;
 75     }
 76 }
 77 
 78 
 79 
 80 void sov(  ) {
 81     for( int x = 1 ; x <= n ; ++x ) {
 82         if( dfn[x] == 0 ) scc( x ) ;
 83     }
 84 }
 85 
 86 
 87 
 88 void output(  ) {
 89     printf( "%d\n" , cnt ) ;    
 90     for( int x = 1 ; x <= cnt ; ++x ) {
 91         printf( "%d " , sum[x] ) ;
 92         for( int y = 0 ; y < sum[x] ; y++ ) 
 93             printf( "%d " , node[x][y] ) ;
 94         printf( "\n" ) ;
 95     }
 96 }
 97 
 98 
 99 int main (  ) {
100 //    Init(  ) ;
101     input(  ) ;
102     sov(  ) ;
103     output(  ) ;
104 //    fclose( stdin ) ;
105 //    fclose( stdout ) ;
106     return 0 ;
107 }