UV 11020, uva11020

UV 11020, uva11020
Ultraviolet A 11020-Efficient Solutions

Question Link

Each person has two attribute values (x, y). For each person (x, y), when there is another person (x', y '), if their property values meet the requirements of X' <x, y' <= y or x' <= x, y' <y, this person will lose the advantage of adding a person each time, and output the current number of advantageous persons

Train of Thought: because each person loses his or her advantage and cannot gain another advantage, he or she can delete these points. In this way, he or she can use a multiset to maintain the point set and add one point each time, use lowerbound and upper_bound to find the position of the vertex next to it for judgment and deletion.

Code:

#include <cstdio>#include <cstring>#include <iostream>#include <set>using namespace std;int t, n;struct Point {    int x, y;    bool operator < (const Point &c) const {if (x == c.x) return y < c.y;return x < c.x;    }};multiset<Point> s;multiset<Point>::iterator it;int main() {    int cas = 0;    cin >> t;    while (t--) {s.clear();cin >> n;Point u;cout << "Case #" << ++cas << ":" << endl;while (n--) {    cin >> u.x >> u.y;    it = s.lower_bound(u);    if (it == s.begin() || (--it)->y > u.y) {s.insert(u);it = s.upper_bound(u);while (it != s.end() && it->y >= u.y) s.erase(it++);    }    cout << s.size() << endl;}if (t) cout << endl;    }    return 0;}