求给定线段集围成的不同闭合区域的对应子集的cpp代码

以下是一个简单的实现，基于扫描线算法和计算几何的知识：

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

const double eps = 1e-8;
const int maxn = 1005;

int dcmp(double x) {
    if (fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}

struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
    bool operator < (const Point& rhs) const {
        return dcmp(x - rhs.x) < 0 || (dcmp(x - rhs.x) == 0 && dcmp(y - rhs.y) < 0);
    }
    bool operator == (const Point& rhs) const {
        return dcmp(x - rhs.x) == 0 && dcmp(y - rhs.y) == 0;
    }
    Point operator + (const Point& rhs) const {
        return Point(x + rhs.x, y + rhs.y);
    }
    Point operator - (const Point& rhs) const {
        return Point(x - rhs.x, y - rhs.y);
    }
    Point operator * (const double& k) const {
        return Point(x * k, y * k);
    }
    double operator * (const Point& rhs) const {
        return x * rhs.y - y * rhs.x;
    }
};

struct Line {
    Point st, ed;
    double angle;
    Line() {}
    Line(Point st, Point ed) : st(st), ed(ed) {
        angle = atan2(ed.y - st.y, ed.x - st.x);
    }
    bool operator < (const Line& rhs) const {
        return angle < rhs.angle;
    }
};

Point get_line_intersection(Line a, Line b) {
    Point u = a.st - b.st;
    double t = (b.ed - b.st) * u / (a.ed - a.st) * (b.ed - b.st);
    return a.st + (a.ed - a.st) * t;
}

bool on_right(Line l, Point p) {
    return dcmp((p - l.st) * (l.ed - l.st)) > 0;
}

vector<Point> HalfPlaneIntersection(vector<Line> lines) {
    vector<Point> p, res;
    sort(lines.begin(), lines.end());
    int n = lines.size();
    for (int i = 0; i < n; i++) {
        if (i && dcmp(lines[i].angle - lines[i - 1].angle) == 0) continue;
        while (p.size() > 1 && on_right(lines[i], p.back())) p.pop_back();
        while (p.size() > 1 && on_right(lines[i], p[p.size() - 2])) p.pop_back();
        p.push_back(get_line_intersection(lines[i], lines[(i + 1) % n]));
    }
    while (p.size() > 1 && on_right(lines[0], p.back())) p.pop_back();
    while (p.size() > 1 && on_right(lines[0], p[p.size() - 2])) p.pop_back();
    if (p.size() < 3) return res;
    for (int i = 0; i < p.size(); i++) {
        res.push_back(p[i]);
    }
    return res;
}

int main() {
    int n;
    cin >> n;
    vector<Line> lines;
    for (int i = 0; i < n; i++) {
        double x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        Point a(x1, y1);
        Point b(x2, y2);
        lines.push_back(Line(a, b));
    }
    vector<Point> res = HalfPlaneIntersection(lines);
    cout << res.size() << endl;
    return 0;
}

其中，`HalfPlaneIntersection` 函数接受一个线段集，返回围成的不同闭合区域的对应点集。