Codeforces round 19D Points(离线处理+Set集合的运用+离散化数据)好题

Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0, 0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:

• add x y — on the sheet of paper Bob marks a point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is not yet marked on Bob's sheet at the time of the request.
• remove x y — on the sheet of paper Bob erases the previously marked point with coordinates (x, y). For each request of this type it's guaranteed that point (x, y) is already marked on Bob's sheet at the time of the request.
• find x y — on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x, y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.

Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2·105, Bob failed to cope. Now he needs a program that will answer all Pete's requests. Help Bob, please!

Input

The first input line contains number n (1 ≤ n ≤ 2·105) — amount of requests. Then there follow n lines — descriptions of the requests. add x y describes the request to add a point, remove x y — the request to erase a point, find x y — the request to find the bottom-left point. All the coordinates in the input file are non-negative and don't exceed 109.

Output

For each request of type find x y output in a separate line the answer to it — coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x, y). If there are no points strictly above and to the right of point (x, y), output -1.

Examples

Input

7
add 1 1
add 3 4
find 0 0
remove 1 1
find 0 0
add 1 1
find 0 0

Output

1 1
3 4
1 1

Input

13
add 5 5
add 5 6
add 5 7
add 6 5
add 6 6
add 6 7
add 7 5
add 7 6
add 7 7
find 6 6
remove 7 7
find 6 6
find 4 4

Output

7 7
-1
5 5

#include<bits/stdc++.h>
using namespace std;

#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)
#define read(x,y) scanf("%d%d",&x,&y)

#define lrt int l,int r,int rt
#define root l,r,rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
/*
题目大意:给定二维平面三种操作，
添加，查询，删除，
对于查询，返回最右上方的最接近的那个点的坐标（先靠左，再靠下的优先级）。

离散+离线+set集合的运用，，这道题很赞。
离散是因为数据范围太大，要记录数据的相对位置，
离线的作用是对于每个数据的相对位置，
我们整体更新它的贡献。

具体的上述操作或多或少都有些小套路在里面。
这道题对于二分查询上下界的基本功要求高，
在离线完后遍历节点数组时，对于每个x，要求在整体相对位置中
找出最小的上界(即线段树查询的下界），
添加和删除都不用说了，线段树是依附在set数组中的，
维护的最大值就是set该位置的最大值，为空则为-1.
查询的话是在给定的边界范围，查询第一个出现比查询值大的那个set集合位置。
那么什么时候可以否定查询呢？线段树当前节点的最大值都没有给定值大，
或者查询区间不满足区间性质（左边界>右边界）,有了这个，就可以
通过线段树查询到最左端满足条件的位置。


*/

const int  maxn =2e5+5;
int n,m;
///离散化数据
struct node
{
    int x,y;
    char op[10];
    void init()
    {
        scanf("%s%d%d",op,&x,&y);
    }
};
node seq[maxn<<2];

set<int> Set[maxn];///线段树建立的基础

int st[maxn<<2],f[maxn];///f是离散化的x坐标映射
void pushup(lrt)
{
    st[rt]=max(st[rt<<1],st[rt<<1|1]);
}
void build(lrt)
{
    st[rt]=-1;
    if(l==r) return ;
    int mid=l+r>>1;
    build(lson),build(rson),pushup(root);
}

void update(lrt,int pos)
{
    if(l==r)
    {
        if(Set[pos].size()) st[rt]=*(--Set[pos].end());
        else st[rt]=-1;
        return ;
    }
    int mid=l+r>>1;
    if(pos<=mid) update(lson,pos);
    if(mid<pos) update(rson,pos);
    pushup(root);
}

int query(int l,int r,int rt,int L,int R,int v)///
{
    if(st[rt]<=v||L>R) return -1;
    if(l==r) return l;
    int mid=l+r>>1;
    if(R<=mid) return query(lson,L,R,v);
    else if(mid<L) return query(rson,L,R,v);
    else
    {
        int q=query(lson,L,R,v);
        if(q==-1) return query(rson,L,R,v);
        else return q;
    }
}

int main()
{
    scanf("%d",&n);for(int i=0;i<n;i++) seq[i].init();

    for(int i=1;i<=n;i++) f[i]=seq[i-1].x;
    sort(f+1,f+1+n);
    m=unique(f+1,f+1+n)-f-1;///压缩后的数据范围
    build(1,m,1);///离散化建立线段树,即每一个相对位置的数据贡献是多少

    for(int i=0;i<n;i++)
    {
        int pos=lower_bound(f+1,f+1+m,seq[i].x)-f;///该坐标的相对位置,起始的查询位置
        if(seq[i].op[0]=='a')
        {
            Set[pos].insert(seq[i].y);
            update(1,m,1,pos);///更新最大值
        }
        else if(seq[i].op[0]=='r')
        {
            Set[pos].erase(seq[i].y);
            update(1,m,1,pos);
        }
        else
        {
            int ans=query(1,m,1,pos+1,m,seq[i].y);///要从线段树中，查询第一个最大值比y大的位置，然后在set集合中那个找其最小的上界
            ///cout<<ans<<" "<<seq[i].y<<endl;
            if(ans==-1) puts("-1");
            else  printf("%d %d\n",f[ans],*Set[ans].upper_bound(seq[i].y));
        }
    }
    return 0;
}