输入两组点云数据，基于DTM方式计算两期土方量中的挖方量的C代码或C++代码的示例有吗

以下是一个基于DTM（数字地形模型）计算挖方量的示例C代码。假设输入的点云数据分别存储在两个数组中，每个数组包含x，y和z坐标的三元组。我们将使用最简单的算法来计算挖方量：对于每个点，将其高程从两期DTM中的较高高程中减去，如果高程差为正则表示该点需要挖掉一定深度的土方。

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

#define MAX_POINTS 100000

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

double distance(struct Point p1, struct Point p2) {
    double dx = p1.x - p2.x;
    double dy = p1.y - p2.y;
    double dz = p1.z - p2.z;
    return sqrt(dx*dx + dy*dy + dz*dz);
}

double compute_volume(struct Point* points1, int num_points1, struct Point* points2, int num_points2) {
    double volume = 0.0;
    int i, j;

    // Compute the bounding box of the two point sets
    double min_x = INFINITY, min_y = INFINITY, min_z = INFINITY;
    double max_x = -INFINITY, max_y = -INFINITY, max_z = -INFINITY;

    for (i = 0; i < num_points1; i++) {
        struct Point p = points1[i];
        if (p.x < min_x) min_x = p.x;
        if (p.y < min_y) min_y = p.y;
        if (p.z < min_z) min_z = p.z;
        if (p.x > max_x) max_x = p.x;
        if (p.y > max_y) max_y = p.y;
        if (p.z > max_z) max_z = p.z;
    }

    for (i = 0; i < num_points2; i++) {
        struct Point p = points2[i];
        if (p.x < min_x) min_x = p.x;
        if (p.y < min_y) min_y = p.y;
        if (p.z < min_z) min_z = p.z;
        if (p.x > max_x) max_x = p.x;
        if (p.y > max_y) max_y = p.y;
        if (p.z > max_z) max_z = p.z;
    }

    // Create the DTM by interpolating between the two point sets
    double dx = (max_x - min_x) / 100.0;
    double dy = (max_y - min_y) / 100.0;
    double** dtm1 = (double**) malloc(101 * sizeof(double*));
    double** dtm2 = (double**) malloc(101 * sizeof(double*));
    for (i = 0; i <= 100; i++) {
        dtm1[i] = (double*) malloc(101 * sizeof(double));
        dtm2[i] = (double*) malloc(101 * sizeof(double));
    }
    for (i = 0; i <= 100; i++) {
        for (j = 0; j <= 100; j++) {
            double x = min_x + i * dx;
            double y = min_y + j * dy;
            int k;
            double