题意：有三种操作“add x y”往平面上添加(x,y)这个点，"remove x y"，将平面上已经存在的点(x,y)删除，“find x y”找出平面上坐标严格大于(x,y)的点，如果有多个点找x最小的，再找y最小的。

思路：将所有点 x 坐标离散化，然后按照新的坐标建一个线段树。对于每一个坐标x，维护一个set，add操作在相应的set里加入y，remove操作在相应的set里减去y。再有一个sum[]数组来维护最值。每次add、remove操作通过pushup操作更新最值。find操作只要找到相应的点即可。

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, n) for (int i = 0; i < n; i++)
#define debug puts("===============")
typedef long long ll;
using namespace std;
int n;
const int maxn = 210000;
int x[maxn], y[maxn], op[maxn], index[maxn], dis[maxn];
char str[11];
int discrete(int x[], int index[], int dis[], int n) {
    int cpy[n];
    memcpy(cpy, x, sizeof(cpy));
    sort(cpy, cpy + n);
    int tot = unique(cpy, cpy + n) - cpy;
    for (int i = 0; i < n; i++) {
        dis[i] = lower_bound(cpy, cpy + tot, x[i]) - cpy;
        index[dis[i]] = x[i];
    }
    return tot;
}
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
set<int> s[maxn];
int mx[maxn << 2];
void build(int l, int r, int rt) {
    mx[rt] = 0;
    if (l == r) {
        s[l].clear();
        return ;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
}
int pushup(int rt) {
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
void add(int pos, int x, int l, int r, int rt) {
    if (l == r) {
        s[l].insert(x);
        mx[rt] = max(mx[rt], x);
        return ;
    }
    int m = (l + r) >> 1;
    if (pos <= m) add(pos, x, lson);
    else add(pos, x, rson);
    pushup(rt);
}
void remove(int pos, int x, int l, int r, int rt) {
    if (l == r) {
        s[l].erase(x);
        mx[rt] = (s[l].empty() ? 0 : *--s[l].end());
        return ;
    }
    int m = (l + r) >> 1;
    if (pos <= m) remove(pos, x, lson);
    else remove(pos, x, rson);
    pushup(rt);
}
pair<int, int> find(int pos, int x, int l, int r, int rt) {
    if (mx[rt] < x || r < pos) return make_pair(l, -1);
    if (l == r) return make_pair(l, *s[l].lower_bound(x));
    int m = (l + r) >> 1;
    pair<int, int> res = find(pos, x, lson);
    if (res.second != -1) return res;
    else return find(pos, x, rson);
}
int main () {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%s%d%d", str, x + i, y + i);
        if (str[0] == 'a') op[i] = 1;
        else if (str[0] == 'r') op[i] = 2;
        else op[i] = 3;
    }
    int m = discrete(x, index, dis, n);
    build(0, m - 1, 1);
    for (int i = 0; i < n; i++) {
        if (op[i] == 1) {
            add(dis[i], y[i], 0, m - 1, 1);
        } else if (op[i] == 2) {
            remove(dis[i], y[i], 0, m - 1, 1);
        } else {
            pair<int, int> res = find(dis[i] + 1, y[i] + 1, 0, m - 1, 1);
            if (res.second == -1) puts("-1");
            else printf("%d %d\n", index[res.first], res.second);
        }
    }
    return 0;
}