输入两组点云数据，分别求其轮廓和其交集的面积的C++代码有吗？

可以的，以下是求解两组点云数据轮廓和交集面积的C代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define MAX_POINTS 100

typedef struct {
    double x;
    double y;
} Point;

typedef struct {
    Point points[MAX_POINTS];
    int count;
} PointSet;

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

double area(PointSet set) {
    double sum = .;
    for (int i = ; i < set.count; i++) {
        int j = (i + 1) % set.count;
        sum += cross(set.points[i], set.points[j]);
    }
    return fabs(sum) / 2.;
}

PointSet convexHull(PointSet set) {
    PointSet hull;
    hull.count = ;
    int n = set.count;
    if (n < 3) {
        for (int i = ; i < n; i++) {
            hull.points[hull.count++] = set.points[i];
        }
        return hull;
    }
    int k = ;
    Point hullPoints[MAX_POINTS];
    for (int i = ; i < n; i++) {
        while (k >= 2 && cross(hullPoints[k - 1] - hullPoints[k - 2], set.points[i] - hullPoints[k - 2]) <= ) {
            k--;
        }
        hullPoints[k++] = set.points[i];
    }
    for (int i = n - 2, t = k + 1; i >= ; i--) {
        while (k >= t && cross(hullPoints[k - 1] - hullPoints[k - 2], set.points[i] - hullPoints[k - 2]) <= ) {
            k--;
        }
        hullPoints[k++] = set.points[i];
    }
    hull.count = k - 1;
    for (int i = ; i < hull.count; i++) {
        hull.points[i] = hullPoints[i];
    }
    return hull;
}

PointSet intersection(PointSet set1, PointSet set2) {
    PointSet inter;
    inter.count = ;
    for (int i = ; i < set1.count; i++) {
        for (int j = ; j < set2.count; j++) {
            Point a = set1.points[i];
            Point b = set1.points[(i + 1) % set1.count];
            Point c = set2.points[j];
            Point d = set2.points[(j + 1) % set2.count];
            double s1 = cross(b - a, c - a);
            double s2 = cross(b - a, d - a);
            double s3 = cross(d - c, a - c);
            double s4 = cross(d - c, b - c);
            if (s1 * s2 <  && s3 * s4 < ) {
                Point p;
                p.x = (s2 * c.x - s1 * d.x) / (s2 - s1);
                p.y = (s2 * c.y - s1 * d.y) / (s2 - s1);
                inter.points[inter.count++] = p;
            }
        }
    }
    return inter;
}

int main() {
    PointSet set1, set2;
    set1.count = 4;
    set1.points[].x = .;
    set1.points[].y = .;
    set1.points[1].x = .;
    set1.points[1].y = 1.;
    set1.points[2].x = 1.;
    set1.points[2].y = 1.;
    set1.points[3].x = 1.;
    set1.points[3].y = .;
    set2.count = 4;
    set2.points[].x = .5;
    set2.points[].y = .5;
    set2.points[1].x = .5;
    set2.points[1].y = 1.5;
    set2.points[2].x = 1.5;
    set2.points[2].y = 1.5;
    set2.points[3].x = 1.5;
    set2.points[3].y = .5;
    PointSet hull1 = convexHull(set1);
    PointSet hull2 = convexHull(set2);
    PointSet inter = intersection(hull1, hull2);
    double area1 = area(hull1);
    double area2 = area(hull2);
    double interArea = area(inter);
    double unionArea = area1 + area2 - interArea;
    printf("轮廓1面积: %lf\n", area1);
    printf("轮廓2面积: %lf\n", area2);
    printf("交集面积: %lf\n", interArea);
    printf("并集面积: %lf\n", unionArea);
    return ;
}