输入两组点云数据（x,y,z），分别求其凸包的轮廓和求两凸包的交集面积的C++代码有吗

对于这个问题，我可以回答。以下是求凸包轮廓和求两凸包交集面积的C代码示例：

求凸包轮廓：

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

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

int cmp(const void *a, const void *b) {
    Point *p1 = (Point *)a;
    Point *p2 = (Point *)b;
    if (p1->x < p2->x) return -1;
    if (p1->x > p2->x) return 1;
    if (p1->y < p2->y) return -1;
    if (p1->y > p2->y) return 1;
    return ;
}

double cross(Point p1, Point p2, Point p3) {
    double x1 = p2.x - p1.x;
    double y1 = p2.y - p1.y;
    double z1 = p2.z - p1.z;
    double x2 = p3.x - p1.x;
    double y2 = p3.y - p1.y;
    double z2 = p3.z - p1.z;
    return x1 * y2 * z3 - x2 * y1 * z3 - x1 * y3 * z2 + x3 * y1 * z2 + x2 * y3 * z1 - x3 * y2 * z1;
}

int convex_hull(Point *points, int n, Point *hull) {
    qsort(points, n, sizeof(Point), cmp);
    int k = ;
    for (int i = ; i < n; i++) {
        while (k >= 2 && cross(hull[k - 2], hull[k - 1], points[i]) <= ) k--;
        hull[k++] = points[i];
    }
    for (int i = n - 2, t = k + 1; i >= ; i--) {
        while (k >= t && cross(hull[k - 2], hull[k - 1], points[i]) <= ) k--;
        hull[k++] = points[i];
    }
    return k - 1;
}

int main() {
    int n;
    scanf("%d", &n);
    Point *points = (Point *)malloc(n * sizeof(Point));
    for (int i = ; i < n; i++) {
        scanf("%lf%lf%lf", &points[i].x, &points[i].y, &points[i].z);
    }
    Point *hull = (Point *)malloc(n * sizeof(Point));
    int m = convex_hull(points, n, hull);
    for (int i = ; i < m; i++) {
        printf("%.2lf %.2lf %.2lf\n", hull[i].x, hull[i].y, hull[i].z);
    }
    free(points);
    free(hull);
    return ;
}

求两凸包交集面积：

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

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

typedef struct {
    Point p1, p2;
} Line;

int cmp(const void *a, const void *b) {
    Point *p1 = (Point *)a;
    Point *p2 = (Point *)b;
    if (p1->x < p2->x) return -1;
    if (p1->x > p2->x) return 1;
    if (p1->y < p2->y) return -1;
    if (p1->y > p2->y) return 1;
    return ;
}

double dot(Point p1, Point p2) {
    return p1.x * p2.x + p1.y * p2.y + p1.z * p2.z;
}

Point cross(Point p1, Point p2) {
    Point p;
    p.x = p1.y * p2.z - p1.z * p2.y;
    p.y = p1.z * p2.x - p1.x * p2.z;
    p.z = p1.x * p2.y - p1.y * p2.x;
    return p;
}

double area(Line l1, Line l2) {
    Point p1 = cross(l1.p2 - l1.p1, l2.p1 - l1.p1);
    Point p2 = cross(l1.p2 - l1.p1, l2.p2 - l1.p1);
    return dot(p1, p2) / 2;
}

int convex_hull(Point *points, int n, Point *hull) {
    qsort(points, n, sizeof(Point), cmp);
    int k = ;
    for (int i = ; i < n; i++) {
        while (k >= 2 && cross(hull[k - 2], hull[k - 1], points[i]) <= ) k--;
        hull[k++] = points[i];
    }
    for (int i = n - 2, t = k + 1; i >= ; i--) {
        while (k >= t && cross(hull[k - 2], hull[k - 1], points[i]) <= ) k--;
        hull[k++] = points[i];
    }
    return k - 1;
}

int main() {
    int n1, n2;
    scanf("%d%d", &n1, &n2);
    Point *points1 = (Point *)malloc(n1 * sizeof(Point));
    for (int i = ; i < n1; i++) {
        scanf("%lf%lf%lf", &points1[i].x, &points1[i].y, &points1[i].z);
    }
    Point *points2 = (Point *)malloc(n2 * sizeof(Point));
    for (int i = ; i < n2; i++) {
        scanf("%lf%lf%lf", &points2[i].x, &points2[i].y, &points2[i].z);
    }
    Point *hull1 = (Point *)malloc(n1 * sizeof(Point));
    int m1 = convex_hull(points1, n1, hull1);
    Point *hull2 = (Point *)malloc(n2 * sizeof(Point));
    int m2 = convex_hull(points2, n2, hull2);
    double area = ;
    for (int i = ; i < m1; i++) {
        for (int j = ; j < m2; j++) {
            Line l1 = {hull1[i], hull1[(i + 1) % m1]};
            Line l2 = {hull2[j], hull2[(j + 1) % m2]};
            area += area(l1, l2);
        }
    }
    printf("%.2lf\n", area);
    free(points1);
    free(points2);
    free(hull1);
    free(hull2);
    return ;
}