spoj 3273 Treap

算是一道比较全面的模板题了吧，需要注意的是：查找比x小的元素个数时x不一定在Treap中，解决办法是插入x，查询，再删除x。

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
using namespace std;

struct Node 
{
    Node * ch[2];
    int v, r, size;
    int cmp( int x )
    {
        if ( x == v ) return -1;
        return x < v ? 0 : 1;
    }    
    void maintain()
    {
        size = 1;
        if ( ch[0] != NULL ) size += ch[0]->size;
        if ( ch[1] != NULL ) size += ch[1]->size;
    }
};

void rotate( Node * & o, int d )
{
    Node * k = o->ch[d ^ 1];
    o->ch[d ^ 1] = k->ch[d];
    k->ch[d] = o;
    o->maintain();
    k->maintain();
    o = k;
}

void insert( Node * & o, int x )
{
    if ( o == NULL )
    {
        o = new Node();
        o->ch[0] = o->ch[1] = NULL;
        o->v = x;
        o->r = rand();
        o->size = 1;
    }
    else
    {
        int d = o->cmp(x);
        insert( o->ch[d], x );
        if ( o->ch[d]->r > o->r )
        {
            rotate( o, d ^ 1 );
        }
        else
        {
            o->maintain();
        }
    }
}

void remove( Node * & o, int x )
{
    int d = o->cmp(x);
    if ( d == -1 )
    {
        if ( o->ch[0] != NULL && o->ch[1] != NULL )
        {
            int dd = ( o->ch[0]->r > o->ch[1]->r ? 1 : 0 );
            rotate( o, dd );
            remove( o->ch[dd], x );
        }
        else
        {
            Node * u = o;
            if ( o->ch[0] == NULL ) o = o->ch[1];
            else o = o->ch[0];
            delete u;
        }
    }
    else
    {
        remove( o->ch[d], x );
    }
    if ( o != NULL ) o->maintain();
}

int ranker( Node * o, int x, int sum )
{
    int d = o->cmp(x);
    if ( d == -1 )
    {
        return sum + ( o->ch[0] == NULL ? 0 : o->ch[0]->size );
    }
    else if ( d == 0 )    
    {
        return ranker( o->ch[0], x, sum );
    }
    else
    {
        int tmp = ( o->ch[0] == NULL ? 0 : o->ch[0]->size );
        return ranker( o->ch[1], x, sum + tmp + 1 );
    }
}

int kth( Node * o, int k )
{
    int tmp = ( o->ch[0] == NULL ? 0 : o->ch[0]->size );
    if ( k == tmp + 1 ) return o->v;
    else if ( k < tmp + 1 ) return kth( o->ch[0], k );
    else return kth( o->ch[1], k - tmp - 1 );
}

int find( Node * o, int x )
{
    if ( o == NULL ) return 0;
    int d = o->cmp(x);
    if ( d == -1 ) return 1;
    return find( o->ch[d], x );
}

int main ()
{
    Node * root = NULL;
    char op[2];
    int q, num;
    scanf("%d", &q);
    while ( q-- )
    {
        scanf("%s%d", op, &num);
        if ( op[0] == 'I' )
        {
            if ( !find( root, num ) )
            {
                insert( root, num );
            }
        }
        else if ( op[0] == 'D' )
        {
            if ( find( root, num ) )
            {
                remove( root, num );
            }
        }
        else if ( op[0] == 'K' )
        {
            int s = ( root == NULL ? 0 : root->size );
            if ( num > s )
            {
                printf("invalid
");
            }
            else
            {
                printf("%d
", kth( root, num ));
            }
        }
        else
        {
            if ( find( root, num ) )
            {
                printf("%d
", ranker( root, num, 0 ));
            }
            else
            {
                insert( root, num );
                printf("%d
", ranker( root, num, 0 ));
                remove( root, num );
            }
        }
    }
    return 0;
}