给我一个计算两个多边形IOU的代码，用C++。

以下是一个计算两个多边形IOU的C++代码，假设多边形的顶点坐标已知：

#include <iostream>
#include <vector>
using namespace std;

struct Point {
    double x, y;
};

// 计算多边形面积
double polygonArea(const vector<Point>& polygon) {
    double area = 0;
    int n = polygon.size();
    for (int i = 0; i < n; i++) {
        int j = (i + 1) % n;
        area += polygon[i].x * polygon[j].y - polygon[j].x * polygon[i].y;
    }
    return area / 2;
}

// 计算多边形的交集面积
double polygonIntersectionArea(const vector<Point>& polygon1, const vector<Point>& polygon2) {
    double area = 0;
    int n1 = polygon1.size();
    int n2 = polygon2.size();
    for (int i = 0; i < n1; i++) {
        int j = (i + 1) % n1;
        for (int k = 0; k < n2; k++) {
            int l = (k + 1) % n2;
            Point p1 = polygon1[i];
            Point p2 = polygon1[j];
            Point q1 = polygon2[k];
            Point q2 = polygon2[l];
            double t1 = (q1.x - p1.x) * (p2.y - p1.y) - (p2.x - p1.x) * (q1.y - p1.y);
            double t2 = (q2.x - p1.x) * (p2.y - p1.y) - (p2.x - p1.x) * (q2.y - p1.y);
            double t3 = (p1.x - q1.x) * (q2.y - q1.y) - (q2.x - q1.x) * (p1.y - q1.y);
            double t4 = (p2.x - q1.x) * (q2.y - q1.y) - (q2.x - q1.x) * (p2.y - q1.y);
            if (t1 * t2 < 0 && t3 * t4 < 0) {
                Point p = {0, 0};
                p.x = (t2 * q1.x - t1 * q2.x) / (t2 - t1);
                p.y = (t2 * q1.y - t1 * q2.y) / (t2 - t1);
                area += polygonArea(vector<Point>{p1, p, p2});
            }
        }
    }
    return area;
}

// 计算两个多边形的IOU
double polygonIOU(const vector<Point>& polygon1, const vector<Point>& polygon2) {
    double area1 = polygonArea(polygon1);
    double area2 = polygonArea(polygon2);
    double intersectionArea = polygonIntersectionArea(polygon1, polygon2);
    double unionArea = area1 + area2 - intersectionArea;
    return intersectionArea / unionArea;
}

int main() {
    // 示例1：边长为1的正方形和边长为2的正方形，重叠面积为1，总面积为4
    vector<Point> polygon1{{0, 0}, {1, 0}, {1, 1}, {0, 1}};
    vector<Point> polygon2{{0.5, 0.5}, {2.5, 0.5}, {2.5, 2.5}, {0.5, 2.5}};
    double iou = polygonIOU(polygon1, polygon2);
    cout << "IOU: " << iou << endl;

    // 示例2：半径为1的圆和半径为2的圆，重叠面积为2.828，总面积为9.424
    const double pi = 3.14159265358979323846;
    vector<Point> circle1;
    vector<Point> circle2;
    int n = 100;
    for (int i = 0; i < n; i++) {
        double angle = 2 * pi / n * i;
        Point p1 = {cos(angle), sin(angle)};
        Point p2 = {2 * cos(angle), 2 * sin(angle)};
        circle1.push_back(p1);
        circle2.push_back(p2);
    }
    iou = polygonIOU(circle1, circle2);
    cout << "IOU: " << iou << endl;

    return 0;
}

以上代码实现了计算两个多边形IOU的功能，并提供了两个示例。其中，`polygonArea`函数用于计算多边形面积；`polygonIntersectionArea`函数用于计算两个多边形的交集面积；`polygonIOU`函数用于计算两个多边形的IOU。