输入两组点云数据，分别求其凸包和交集的面积的C++代码有吗？

是的，以下是使用C++实现求两组点云数据的凸包和交集面积的代码：

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

struct Point {
    double x, y;

    Point() {}
    Point(double x, double y) : x(x), y(y) {}

    bool operator<(const Point& p) const {
        return x < p.x || (x == p.x && y < p.y);
    }

    double distance(const Point& p) const {
        double dx = x - p.x;
        double dy = y - p.y;
        return sqrt(dx * dx + dy * dy);
    }

    double cross(const Point& p) const {
        return x * p.y - y * p.x;
    }

    Point operator-(const Point& p) const {
        return Point(x - p.x, y - p.y);
    }
};

double cross(const Point& O, const Point& A, const Point& B) {
    return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
}

vector<Point> convex_hull(vector<Point>& points) {
    int n = points.size(), k = 0;
    vector<Point> hull(2 * n);

    sort(points.begin(), points.end());

    for (int i = 0; i < n; i++) {
        while (k >= 2 && cross(hull[k - 2], hull[k - 1], points[i]) <= 0) k--;
        hull[k++] = points[i];
    }

    for (int i = n - 2, t = k + 1; i >= 0; i--) {
        while (k >= t && cross(hull[k - 2], hull[k - 1], points[i]) <= 0) k--;
        hull[k++] = points[i];
    }

    hull.resize(k - 1);
    return hull;
}

double polygon_area(vector<Point>& points) {
    int n = points.size();
    double area = 0.0;

    for (int i = 1; i < n - 1; i++) {
        double x1 = points[i].x - points[0].x;
        double y1 = points[i].y - points[0].y;
        double x2 = points[i + 1].x - points[0].x;
        double y2 = points[i + 1].y - points[0].y;
        area += x1 * y2 - x2 * y1;
    }

    return abs(area) / 2.0;
}

vector<Point> intersection(vector<Point>& a, vector<Point>& b) {
    vector<Point> res;

    for (int i = 0; i < a.size(); i++) {
        Point p1 = a[i];
        Point p2 = a[(i + 1) % a.size()];

        for (int j = 0; j < b.size(); j++) {
            Point q1 = b[j];
            Point q2 = b[(j + 1) % b.size()];

            double d1 = cross(p1, p2, q1);
            double d2 = cross(p1, p2, q2);
            double d3 = cross(q1, q2, p1);
            double d4 = cross(q1, q2, p2);

            if ((d1 > 0 && d2 < 0 || d1 < 0 && d2 > 0) &&
                (d3 > 0 && d4 < 0 || d3 < 0 && d4 > 0)) {

                double t = d1 / (d1 - d2);
                double x = q1.x + t * (q2.x - q1.x);
                double y = q1.y + t * (q2.y - q1.y);
                res.push_back(Point(x, y));
            }
        }
    }

    return convex_hull(res);
}

int main() {
    vector<Point> a = { {0.0, 0.0}, {0.0, 1.0}, {1.0, 0.0}, {1.0, 1.0} };
    vector<Point> b = { {0.5, 0.5}, {0.5, 1.5}, {1.5, 0.5}, {1.5, 1.5} };

    vector<Point> hull_a = convex_hull(a);
    vector<Point> hull_b = convex_hull(b);

    double area_a = polygon_area(hull_a);
    double area_b = polygon_area(hull_b);

    vector<Point> inter = intersection(hull_a, hull_b);
    double area_inter = polygon_area(inter);

    cout << "Convex hull of A: ";
    for (auto& p : hull_a) {
        cout << "(" << p.x << ", " << p.y << ") ";
    }
    cout << endl;

    cout << "Convex hull of B: ";
    for (auto& p : hull_b) {
        cout << "(" << p.x << ", " << p.y << ") ";
    }
    cout << endl;

    cout << "Intersection of A and B: ";
    for (auto& p : inter) {
        cout << "(" << p.x << ", " << p.y << ") ";
    }
    cout << endl;

    cout << "Area of A: " << area_a << endl;
    cout << "Area of B: " << area_b << endl;
    cout << "Area of intersection: " << area_inter << endl;

    return 0;
}

在这个例子中，我们使用了四个点构成两组点云数据A和B，然后使用convex_hull函数求出了它们的凸包，接着使用polygon_area函数计算凸包面积，最后使用intersection函数求出两组点云数据的交集，并使用polygon_area函数计算交集面积。