求强连通分量

还是贴一份tarjan的代码吧 ，QwQ

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
using namespace std ;
const int inf = 1 << 30 , maxn = 10000 + 100 , maxe = 500000 + 100 ;
int  n , m , cnt ,head[maxn] , tot  , top , sta[maxn], dfn[maxn] , low[maxn] , sum[maxn];
struct id {
    int to , nxt ;
}  edge[maxe] ;
bool lie[maxe] ;
vector< vector<int> > node( maxn ) ; 


inline void Init(  ) {
    freopen( "10717.in" , "r" , stdin ) ;
    freopen( "10717.out" , "w" , stdout ) ;
}

int read(  ) {
    char ch = getchar(  ) ; int ret = 0 , k = 1 ;
    while( ch < '0' || ch > '9' ) { if( ch == '-' ) k = -1 ; ch = getchar(  ) ; }
    while( ch >= '0' && ch <= '9' ) ret = ret * 10 + ch - '0' , ch = getchar(  ) ;
    return ret * k ;
}


void add( int u , int v ) {
    edge[++cnt].to = v ;
    edge[cnt].nxt = head[u] ;
    head[u] = cnt ;
}


void input(  ) {
    n = read(  ) , m = read(  ) ;
    int  u , v ;
    memset( head , -1 , sizeof( head ) ) ;
    for( int x = 1 ;  x <= m ; ++x ) {
        u = read( ) , v = read(  ) ;
        add( u , v ) ;
    }
    cnt = 0 ;
}


void scc( int u ) {
    dfn[u] = ++tot ; low[u] = tot ;
    sta[++top] = u ; lie[u] = true ;
    for( int i = head[u] ; ~i ; i = edge[i].nxt ) {
        int v = edge[i].to ;
        if( dfn[v] == 0 ) 
        {
            scc( v ) ;
            low[u] = min( low[u] , low[v] ) ;
        }
        else if( lie[v] ) {
            low[u] = min( low[u] , dfn[v] ) ;
        }    
    }
    if( low[u] == dfn[u] ) {
        cnt ++ ;
        int front = 0 ;
        do {
            front = sta[top--] ;
            sum[cnt] ++ ;
            node[cnt].push_back( front ) ;
            lie[front] = false ;
            
        } while( front != u ) ;
    }
}



void sov(  ) {
    for( int x = 1 ; x <= n ; ++x ) {
        if( dfn[x] == 0 ) scc( x ) ;
    }
}



void output(  ) {
    printf( "%d
" , cnt ) ;    
    for( int x = 1 ; x <= cnt ; ++x ) {
        printf( "%d " , sum[x] ) ;
        for( int y = 0 ; y < sum[x] ; y++ ) 
            printf( "%d " , node[x][y] ) ;
        printf( "
" ) ;
    }
}


int main (  ) {
//    Init(  ) ;
    input(  ) ;
    sov(  ) ;
    output(  ) ;
//    fclose( stdin ) ;
//    fclose( stdout ) ;
    return 0 ;
}