题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3237

对询问序列进行分治，然后利用并查集进行缩点（修改边即可），这样时间复杂度就可以达到了O(k log k)。

代码：

#include <cstdio>

#include <algorithm>

#include <cstring>

  

using namespace std ;

  

#define DOWN( i , r , l ) for ( int i = r ; i >= l ; -- i )

#define rep( i , x ) for ( int i = 0 ; i ++ < x ; )

#define REP( i , l , r ) for ( int i = l ; i <= r ; ++ i )

#define Rep( i , x ) for ( int i = 0 ; i < x ; ++ i )

  

const int maxn = 101000 , maxm = 201000 , maxk = 101000 , maxd = 19 ;

  

int father[ maxn ] ;

  

inline void Init_us( int V ) {

        rep( i , V ) father[ i ] = 0 ;

}

  

inline int getfa( int now ) {

        int i = now ; for ( ; father[ i ] ; i = father[ i ] ) ;

        for ( int j = now , k ; father[ j ] ; k = father[ j ] , father[ j ] = i , j = k ) ;

        return i ;

}

  

inline void merge( int s , int t ) {

        int u = getfa( s ) , v = getfa( t ) ;

        if ( u != v ) father[ u ] = v ;

}

  

int edge[ maxm ][ 2 ] , en[ maxk ] , e[ maxk ][ 4 ][ 3 ] , n , m , k , num[ maxn ] , temp[ maxd ][ maxk ][ 4 ][ 2 ] , trans[ maxn ] ;

bool used[ maxm ] , ans[ maxk ] ;

  

inline int transform( int V , int l , int r ) {

        rep( i , V ) num[ i ] = 0 ;

        int cnt = 0 , now ;

        rep( i , V ) {

                if ( ! num[ now = getfa( i ) ] ) num[ now ] = ++ cnt ;

                trans[ i ] = num[ now ] ;

        }

        REP( i , l , r ) Rep( j , en[ i ] ) {

                e[ i ][ j ][ 0 ] = trans[ e[ i ][ j ][ 0 ] ] , e[ i ][ j ][ 1 ] = trans[ e[ i ][ j ][ 1 ] ] ;

        }

        return cnt ;

}

  

inline void Save( int l , int r , int lev ) {

        REP( i , l , r ) Rep( j , en[ i ] ) {

                temp[ lev ][ i ][ j ][ 0 ] = e[ i ][ j ][ 0 ] , temp[ lev ][ i ][ j ][ 1 ] = e[ i ][ j ][ 1 ] ;

        }

}

  

inline void Load( int l , int r , int lev ) {

        REP( i , l , r ) Rep( j , en[ i ] ) {

                e[ i ][ j ][ 0 ] = temp[ lev ][ i ][ j ][ 0 ] , e[ i ][ j ][ 1 ] = temp[ lev ][ i ][ j ][ 1 ] ;

        }

}

  

int VI ;

  

inline void Init(  ) {

        scanf( "%d%d" , &n , &m ) ;

        rep( i , m ) scanf( "%d%d" , edge[ i ] , edge[ i ] + 1 ) ;

        scanf( "%d" , &k ) ;

        memset( used , false , sizeof( used ) ) ;

        rep( i , k ) {

                scanf( "%d" , en + i ) ;

                Rep( j , en[ i ] ) {

                        int x ; scanf( "%d" , &x ) ; used[ x ] = true ;

                        e[ i ][ j ][ 0 ] = edge[ x ][ 0 ] , e[ i ][ j ][ 1 ] = edge[ x ][ 1 ] , e[ i ][ j ][ 2 ] = x ;

                }

        }

        Init_us( n ) ;

        rep( i , m ) if ( ! used[ i ] ) merge( edge[ i ][ 0 ] , edge[ i ][ 1 ] ) ;

        VI = transform( n , 1 , k ) ;

}

  

int flag[ maxm ] , tag = 0 ;

  

void solve( int V , int l , int r , int lev ) {

        if ( l == r ) {

                ans[ l ] = V == 1 ; return ;

        }

        int mid = ( l + r ) >> 1 ;

        ++ tag ;

        REP( i , l , mid ) Rep( j , en[ i ] ) flag[ e[ i ][ j ][ 2 ] ] = tag ;

        Init_us( V ) ;

        REP( i , ( mid + 1 ) , r ) Rep( j , en[ i ] ) if ( flag[ e[ i ][ j ][ 2 ] ] < tag ) {

                merge( e[ i ][ j ][ 0 ] , e[ i ][ j ][ 1 ] ) ;

        }

        Save( l , mid , lev ) ;

        int v = transform( V , l , mid ) ;

        solve( v , l , mid , lev + 1 ) ;

        Load( l , mid , lev ) ;

        ++ tag ;

        REP( i , ( mid + 1 ) , r ) Rep( j , en[ i ] ) flag[ e[ i ][ j ][ 2 ] ] = tag ;

        Init_us( V ) ;

        REP( i , l , mid ) Rep( j , en[ i ] ) if ( flag[ e[ i ][ j ][ 2 ] ] < tag ) {

                merge( e[ i ][ j ][ 0 ] , e[ i ][ j ][ 1 ] ) ;

        }

        v = transform( V , mid + 1 , r ) ;

        solve( v , mid + 1 , r , lev + 1 ) ;

}

  

int main(  ) {

        Init(  ) ;

        memset( flag , 0 , sizeof( flag ) ) ;

        solve( VI , 1 , k , 0 ) ;

        rep( i , k ) printf( ans[ i ] ? "Connected\n" : "Disconnected\n" ) ;

        return 0 ;

}