输入两组点云数据进行快速凸包构建，求其交集面积有没有具体的C或C++代码示例

可以的，以下是一个基于C++的快速凸包算法示例代码：

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

using namespace std;

struct Point {
    double x, y;
};

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

double area(const vector<Point>& p) {
    double res = ;
    for (int i = ; i < p.size(); i++) {
        int j = (i + 1) % p.size();
        res += cross(p[i], p[j]);
    }
    return res / 2;
}

vector<Point> convex_hull(vector<Point> p) {
    sort(p.begin(), p.end(), [](const Point& a, const Point& b) {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    });

    vector<Point> h;
    for (int i = ; i < p.size(); i++) {
        while (h.size() >= 2 && cross(h.back() - h[h.size() - 2], p[i] - h.back()) <= ) {
            h.pop_back();
        }
        h.push_back(p[i]);
    }

    int t = h.size();
    for (int i = p.size() - 2; i >= ; i--) {
        while (h.size() > t && cross(h.back() - h[h.size() - 2], p[i] - h.back()) <= ) {
            h.pop_back();
        }
        h.push_back(p[i]);
    }

    h.pop_back();
    return h;
}

vector<Point> intersection(const vector<Point>& p1, const vector<Point>& p2) {
    vector<Point> res;

    for (int i = ; i < p1.size(); i++) {
        int j = (i + 1) % p1.size();
        double a1 = p1[j].y - p1[i].y;
        double b1 = p1[i].x - p1[j].x;
        double c1 = cross(p1[i], p1[j]);
        for (int k = ; k < p2.size(); k++) {
            int l = (k + 1) % p2.size();
            double a2 = p2[l].y - p2[k].y;
            double b2 = p2[k].x - p2[l].x;
            double c2 = cross(p2[k], p2[l]);
            double det = a1 * b2 - a2 * b1;
            if (fabs(det) < 1e-9) {
                continue;
            }
            double x = (b2 * c1 - b1 * c2) / det;
            double y = (a1 * c2 - a2 * c1) / det;
            Point p = {x, y};
            bool inside1 = true, inside2 = true;
            for (int m = ; m < p1.size(); m++) {
                if (cross(p1[m] - p, p1[(m + 1) % p1.size()] - p) < ) {
                    inside1 = false;
                    break;
                }
            }
            for (int m = ; m < p2.size(); m++) {
                if (cross(p2[m] - p, p2[(m + 1) % p2.size()] - p) < ) {
                    inside2 = false;
                    break;
                }
            }
            if (inside1 && inside2) {
                res.push_back(p);
            }
        }
    }

    return convex_hull(res);
}

int main() {
    vector<Point> p1 = {{, }, {1, }, {1, 1}, {, 1}};
    vector<Point> p2 = {{.5, .5}, {1.5, .5}, {1.5, 1.5}, {.5, 1.5}};
    vector<Point> p = intersection(p1, p2);
    cout << area(p) << endl;
    return ;
}

这个示例代码实现了快速凸包算法和求两个凸包的交集面积。你可以将两组点云数据分别存储在两个vector<Point>中，然后调用intersection函数求它们的交集，最后调用area函数计算交集面积。