Cod sursa(job #1220148)

Utilizator lucian666Vasilut Lucian lucian666 Data 16 august 2014 17:14:25
Problema Lowest Common Ancestor Scor 0
Compilator cpp Status done
Runda Arhiva educationala Marime 2.09 kb



#include <fstream>
#include <vector>
#include <algorithm>
#include <cmath>

#define NMAX 100009
#define LGMAX 20
#define pb push_back

using namespace std;
ofstream out("lca.out");
ifstream in("lca.in");

int n , m , K , H[NMAX] , L[NMAX] , First[NMAX] , dp[NMAX][LGMAX] ;

vector <int>G[NMAX];
typedef vector<int>:: iterator IT;

void read();
void dfs( int nod , int level );
void RMQ();
void wrs();

int main()
{

    read();
    dfs(1,0);
    RMQ();
    wrs();

    return 0;

}

void read()
{

    in >> n >> m;
    for( int i=2 ; i<=n ; i++)
    {
        int x;
        in >> x;
        G[x].pb(i);
    }


}

void dfs( int nod , int level )
{

    H[ ++K ] = nod;
    L[ K ] = level;
    First[ nod ] = K;

    for( IT i = G[nod].begin() ; i!= G[nod].end(); ++i )
    {

        dfs( *i , level + 1 );
        H[ ++K ] = nod;
        L[ K ] = level;

    }

}

void RMQ()
{
    /* ca sa fie normal tre sa modific :
        for( int i=1 ; i<=K ; ++i )
        dp[i][0] = H[i];

        iar in dinamica ....
        if( i + ( 1 << ( j - 1 )) <=K )
                dp[i][j] = min( dp[i][j] ,  dp[ i + ( 1 << ( j-1 ) ) ][j-1] ) ;

    */
    //rmq de aici imi da poz minimului din vectorul L
    for( int i=1 ; i<=K ; ++i )
        dp[i][0] = i;


    int k = 0;
    while( ( 1 << k ) <=K )
        ++k;

    --k;

    for( int j=1 ; j<=k ; ++j )
    {

        for( int i=1 ; i<=K ; ++i )
        {

            dp[i][j] = dp[i][j-1];

            if( i + ( 1 << ( j - 1 )) <=K )
                if( L [ dp[i][j] ] > L [ dp[ i + ( 1 << j -1 ) ][j-1] ] )
                dp[i][j] =  dp[ i + ( 1 << ( j-1 ) ) ][j-1] ;

        }
    }



}


void wrs()
{

    for( int i , j ; m ; --m )
    {
        in >> i >> j;
       // out << i << " " << j << '\n';

        int x , y;

        x = First[i];
        y = First[j];

        if( x > y )
            swap(x,y);

        int k = floor( log ( y - x ) / log( 2 ));

        out << H[ min ( dp[x][k] , dp[ y - ( 1 << k ) + 1][k]  ) ]<< '\n';


    }

}