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

可以回答这个问题。以下是 C 语言的代码示例：

求凸包轮廓：

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

#define MAX_POINTS 100

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

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

int orientation(Point p, Point q, Point r) {
double val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
if (val == )
return ;
return (val > ) ? 1 : 2;
}

void convexHull(Point points[], int n, Point hull[], int *hullSize) {
qsort(points, n, sizeof(Point), cmp);
int i, j, k = ;
Point temp[MAX_POINTS * 2];
for (i = ; i < n; i++) {
while (k >= 2 && orientation(temp[k - 2], temp[k - 1], points[i]) != 2)
k--;
temp[k++] = points[i];
}
for (i = n - 2, j = k + 1; i >= ; i--) {
while (k >= j && orientation(temp[k - 2], temp[k - 1], points[i]) != 2)
k--;
temp[k++] = points[i];
}
*hullSize = k - 1;
for (i = ; i < *hullSize; i++)
hull[i] = temp[i];
}

求两凸包的交集面积：

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

#define MAX_POINTS 100

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

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

int orientation(Point p, Point q, Point r) {
double val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);
if (val == )
return ;
return (val > ) ? 1 : 2;
}

void convexHull(Point points[], int n, Point hull[], int *hullSize) {
qsort(points, n, sizeof(Point), cmp);
int i, j, k = ;
Point temp[MAX_POINTS * 2];
for (i = ; i < n; i++) {
while (k >= 2 && orientation(temp[k - 2], temp[k - 1], points[i]) != 2)
k--;
temp[k++] = points[i];
}
for (i = n - 2, j = k + 1; i >= ; i--) {
while (k >= j && orientation(temp[k - 2], temp[k - 1], points[i]) != 2)
k--;
temp[k++] = points[i];
}
*hullSize = k - 1;
for (i = ; i < *hullSize; i++)
hull[i] = temp[i];
}

double area(Point p[], int n) {
double area = ;
int i;
for (i = ; i < n; i++) {
int j = (i + 1) % n;
area += p[i].x * p[j].y - p[j].x * p[i].y;
}
return fabs(area) / 2;
}

double intersectionArea(Point p1[], int n1, Point p2[], int n2) {
Point hull1[MAX_POINTS], hull2[MAX_POINTS];
int hullSize1, hullSize2;
convexHull(p1, n1, hull1, &hullSize1);
convexHull(p2, n2, hull2, &hullSize2);
double area1 = area(hull1, hullSize1);
double area2 = area(hull2, hullSize2);
int i, j, k, l;
double area = ;
for (i = ; i < hullSize1; i++) {
j = (i + 1) % hullSize1;
for (k = ; k < hullSize2; k++) {
l = (k + 1) % hullSize2;
double x1 = hull1[i].x, y1 = hull1[i].y;
double x2 = hull1[j].x, y2 = hull1[j].y;
double x3 = hull2[k].x, y3 = hull2[k].y;
double x4 = hull2[l].x, y4 = hull2[l].y;
double a1 = y2 - y1, b1 = x1 - x2, c1 = x1 * y2 - x2 * y1;
double a2 = y4 - y3, b2 = x3 - x4, c2 = x3 * y4 - x4 * y3;
double det = a1 * b2 - a2 * b1;
if (det == )
continue;
double x = (b1 * c2 - b2 * c1) / det;
double y = (a2 * c1 - a1 * c2) / det;
if (x < fmax(x1, x2) || x > fmin(x1, x2) || y < fmax(y1, y2) || y > fmin(y1, y2))
continue;
if (x < fmax(x3, x4) || x > fmin(x3, x4) || y < fmax(y3, y4) || y > fmin(y3, y4))
continue;
Point p = {x, y, };
Point temp1[MAX_POINTS], temp2[MAX_POINTS];
int tempSize1 = , tempSize2 = ;
for (int m = ; m < hullSize1; m++) {
int n = (m + 1) % hullSize1;
if (orientation(hull1[m], hull1[n], p) != 2)
temp1[tempSize1++] = hull1[m];
}
for (int m = ; m < hullSize2; m++) {
int n = (m + 1) % hullSize2;
if (orientation(hull2[m], hull2[n], p) != 2)
temp2[tempSize2++] = hull2[m];
}
temp1[tempSize1++] = p;
temp2[tempSize2++] = p;
area += area(temp1, tempSize1) + area(temp2, tempSize2);
}
}
return area1 + area2 - area / 2;
}