对点云数据求凸包的最小外接矩形的c++代码

以下是对点云数据求凸包的最小外接矩形的C++代码：

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

using namespace std;

// 定义点结构体
struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
};

// 定义向量结构体
struct Vector {
    double x, y;
    Vector() {}
    Vector(double x, double y) : x(x), y(y) {}
    Vector(Point A, Point B) : x(B.x - A.x), y(B.y - A.y) {}
    double length() { return hypot(x, y); } // 计算向量长度
};

// 两个点是否相等
bool operator==(const Point& A, const Point& B) {
    return A.x == B.x && A.y == B.y;
}

// 向量点乘
double operator*(const Vector& A, const Vector& B) {
    return A.x * B.x + A.y * B.y;
}

// 向量叉乘
double operator^(const Vector& A, const Vector& B) {
    return A.x * B.y - A.y * B.x;
}

// 求点集的凸包
vector<Point> convexHull(vector<Point> points) {
    int n = points.size();
    if (n < 3) return points;
    sort(points.begin(), points.end(), [](const Point& A, const Point& B) {
        return A.x < B.x || (A.x == B.x && A.y < B.y);
    });
    vector<Point> ch;
    for (int i = 0; i < n; i++) {
        while (ch.size() >= 2 && ((ch[ch.size() - 1] - ch[ch.size() - 2]) ^ (points[i] - ch[ch.size() - 2])) <= 0) {
            ch.pop_back();
        }
        ch.push_back(points[i]);
    }
    int k = ch.size();
    for (int i = n - 2; i >= 0; i--) {
        while (ch.size() >= k + 1 && ((ch[ch.size() - 1] - ch[ch.size() - 2]) ^ (points[i] - ch[ch.size() - 2])) <= 0) {
            ch.pop_back();
        }
        ch.push_back(points[i]);
    }
    if (ch.size() > 1) ch.pop_back();
    return ch;
}

// 求点集的最小外接矩形
double minBoundingRect(vector<Point> points) {
    vector<Point> ch = convexHull(points);
    int n = ch.size();
    if (n < 3) return 0;
    double ans = 1e18;
    for (int i = 0, j = 2; i < n; i++) {
        Vector v1(ch[i], ch[(i + 1) % n]), v2;
        while ((v2 = Vector(ch[i], ch[(j + 1) % n])) ^ v1 >= 0) j = (j + 1) % n;
        double len1 = v1.length(), len2 = v2.length();
        ans = min(ans, len1 * len2 * abs(v1 * v2) / 2 / abs(v1 ^ v2));
    }
    return ans;
}

int main() {
    // 输入点云数据
    vector<Point> points = {{-1, 0}, {0, 2}, {2, 1}, {1, -1}};

    // 求点集的最小外接矩形
    double rect = minBoundingRect(points);
    cout << "Minimum bounding rectangle: " << rect << endl;

    return 0;
}

注意：上述代码中的 `minBoundingRect` 函数需要先求出点集的凸包，然后枚举凸包上的每条边，计算以该边为对角线的矩形面积，并取所有面积的最小值。