学步园

传送门：【CodeForces】19D Points

题目分析：将X轴离散化，然后构造一棵线段树，线段树的每个叶子节点维护一个set，里面为坐标为a[x]的点的y坐标的集合。线段树维护区间内所有点能达到的y的最大值。

每次插入删除就单点更新，询问的时候看区间【x+1，cnt】内maxv[ o ]是否大于y，不存在返回0，否则返回最左边满足集合内存在y坐标大于查询的值的下标。然后在下标所在的集合查询最小的大于y的y'值即可。

set真是强大！

代码如下：

#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )
#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define mid ( ( l + r ) >> 1 )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 1 , cnt
#define rt o , l , r
#define mid ( ( l + r ) >> 1 )

const int MAXN = 200005 ;

struct Node {
	int x , y ;
	char op ;
} opp[MAXN] ;

int maxv[MAXN << 2] ;
int a[MAXN] , cnt ;
set < int > p[MAXN] ;
int n ;

void update ( int pos , int o , int l , int r ) {
	if ( l == r ) {
		if ( p[pos].empty () ) maxv[o] = 0 ;
		else maxv[o] = *p[pos].rbegin () ;//返回end的值
		return ;
	}
	int m = mid ;
	if ( pos <= m ) update ( pos , lson ) ;
	else            update ( pos , rson ) ;
	maxv[o] = max ( maxv[ls] , maxv[rs] ) ;
}

int query ( int L , int R , int key , int o , int l , int r ) {
	if ( maxv[o] <= key ) return 0 ;
	if ( l == r ) return l ;
	int m = mid ;
	if ( R <= m ) return query ( L , R , key , lson ) ;
	if ( m <  L ) return query ( L , R , key , rson ) ;
	int res = query ( L , R , key , lson ) ;
	if ( !res ) res = query ( L , R , key , rson ) ;
	return res ;
}

int unique ( int n ) {
	int cnt = 1 ;
	sort ( a + 1 , a + n + 1 ) ;
	FOR ( i , 2 , n ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ;
	return cnt ;
}

int hash ( int x ) {
	int l = 1 , r = cnt , m ;
	while ( l < r ) {
		m = mid ;
		if ( a[m] >= x ) r = m ;
		else l = m + 1 ;
	}
	return l ;
}

void solve () {
	char buf[10] ;
	FOR ( i , 1 , n ) {
		scanf ( "%s%d%d" , buf , &opp[i].x , &opp[i].y ) ;
		a[i] = opp[i].x ;
		opp[i].op = buf[0] ;
		p[i].clear () ;
	}
	cnt = unique ( n ) ;
	FOR ( i , 1 , n ) {
		int x = hash ( opp[i].x ) , y = opp[i].y ;
		if ( opp[i].op == 'a' ) {
			p[x].insert ( y ) ;
			update ( x , root ) ;
		} else if ( opp[i].op == 'r' ) {
			p[x].erase ( y ) ;
			update ( x , root ) ;
		} else {
			if ( x + 1 > cnt ) x = 0 ;
			if ( x ) x = query ( x + 1 , cnt , y , root ) ;
			if ( x ) printf ( "%d %d\n" , a[x] , *p[x].upper_bound ( y ) ) ;
			else printf ( "-1\n" ) ;
		}
	}
}

int main () {
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}