题目链接：
https://cn.vjudge.net/problem/CodeForces-19D
题目大意：
n个操作，在200000*200000的平面上加删点
find 严格在坐标右上角，x最小，再y最小的点
解题思路：
线段树，离散化x坐标，线段树中保存y最大值，这样可以找到严格大于点x' y'的最小的x，用set存储每个x的y，就可以找到大于y'的y
1 #include<bits/stdc++.h> 2 #define mid(l, r) ((l) + ((r) - (l)) / 2) 3 #define lc ((o)<<1) 4 #define rc ((o)<<1|1) 5 using namespace std; 6 const int maxn = 1e6 + 10; 7 typedef long long ll; 8 struct Edge 9 { 10 int op, x, y; 11 Edge(){} 12 Edge(int op, int x, int y):op(op), x(x), y(y){} 13 }a[maxn];//存储离线操作 14 int num[maxn];//去重 15 struct node 16 { 17 int l, r, y;//每个x存储最大的y 18 }tree[maxn];//线段树 19 set<int>tot[maxn];//存储x中的y 20 21 void build(int o, int l, int r) 22 { 23 tree[o].l = l, tree[o].r = r, tree[o].y = -1; 24 if(l == r)return; 25 int m = mid(l, r); 26 build(lc, l, m); 27 build(rc, m + 1, r); 28 } 29 int p; 30 void update(int o) 31 { 32 if(tree[o].l == tree[o].r) 33 { 34 if(tot[p].size())tree[o].y = *(--tot[p].end()); 35 else tree[o].y = -1; 36 return; 37 } 38 if(p <= tree[lc].r)update(lc); 39 else update(rc); 40 tree[o].y = max(tree[lc].y, tree[rc].y); 41 } 42 int query(int x, int y, int o) 43 { 44 if(tree[o].r <= x)return -1; 45 if(tree[o].y <= y)return -1; 46 if(tree[o].l == tree[o].r)return tree[o].l; 47 int t = query(x, y, lc); 48 if(t == -1)t = query(x, y, rc); 49 return t; 50 } 51 int main() 52 { 53 int n;char s[10]; 54 cin >> n; 55 for(int i = 1; i <= n; i++) 56 { 57 scanf("%s%d%d", s, &a[i].x, &a[i].y); 58 if(s[0] == 'a')a[i].op = 1; 59 else if(s[0] == 'r')a[i].op = 2; 60 else a[i].op = 3; 61 num[i] = a[i].x; 62 } 63 sort(num + 1, num + 1 + n); 64 int m = unique(num + 1, num + 1 + n) - (num + 1); 65 build(1, 1, m); 66 for(int i = 1; i <= n; i++) 67 { 68 int x = upper_bound(num + 1, num + 1 + m, a[i].x) - (num + 1); 69 int y = a[i].y; 70 p = x; 71 if(a[i].op == 1) 72 { 73 tot[x].insert(y); 74 update(1); 75 } 76 else if(a[i].op == 2) 77 { 78 tot[x].erase(y); 79 update(1); 80 } 81 else 82 { 83 int ans = query(x, y, 1); 84 if(ans == -1)puts("-1"); 85 else printf("%d %d ", num[ans], *tot[ans].upper_bound(y)); 86 } 87 } 88 return 0; 89 }