请帮我写一个水准网间接平差的c++代码

以下是一个简单的水准网间接平差的C++代码，仅供参考：

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;

// 定义一个结构体表示测站坐标和高程
struct Station {
    double x;
    double y;
    double z;
};

// 定义一个结构体表示测站之间的连接关系和观测量
struct Observation {
    int from;
    int to;
    double height_diff; // 水准高差
    double horiz_angle; // 水平角
    double azimuth; // 方位角
};

// 定义一个函数，根据观测量计算各测站的相对坐标和高程差
vector<Station> indirectAdjustment(const vector<Station>& stations, const vector<Observation>& observations) {
    int n = stations.size();
    vector<vector<double>> A(3 * (n - 1), vector<double>(3 * n, 0)); // 系数矩阵
    vector<double> b(3 * (n - 1), 0); // 常数向量

    // 构建系数矩阵和常数向量
    for (int i = 0; i < observations.size(); i++) {
        int from = observations[i].from;
        int to = observations[i].to;
        double height_diff = observations[i].height_diff;
        double horiz_angle = observations[i].horiz_angle;
        double azimuth = observations[i].azimuth;

        double cos_h = cos(horiz_angle);
        double sin_h = sin(horiz_angle);
        double cos_a = cos(azimuth);
        double sin_a = sin(azimuth);

        // 水准高差方程
        A[3 * i][3 * from] = -1;
        A[3 * i][3 * to] = 1;
        b[3 * i] = height_diff;

        // 坐标方程
        A[3 * i + 1][3 * from] = -cos_h * sin_a;
        A[3 * i + 1][3 * from + 1] = -cos_h * cos_a;
        A[3 * i + 1][3 * from + 2] = sin_h;
        A[3 * i + 1][3 * to] = cos_h * sin_a;
        A[3 * i + 1][3 * to + 1] = cos_h * cos_a;
        A[3 * i + 1][3 * to + 2] = -sin_h;

        // 高程方程
        A[3 * i + 2][3 * from + 2] = -1;
        A[3 * i + 2][3 * to + 2] = 1;
        b[3 * i + 2] = stations[from].z - stations[to].z;
    }

    // 求解方程 Ax=b
    vector<vector<double>> AtA(3 * n, vector<double>(3 * n, 0)); // 系数矩阵的转置矩阵乘以系数矩阵
    vector<double> Atb(3 * n, 0); // 系数矩阵的转置矩阵乘以常数向量
    for (int i = 0; i < 3 * (n - 1); i++) {
        for (int j = 0; j < 3 * n; j++) {
            AtA[j][i] = A[i][j];
        }
        Atb[i] = b[i];
    }
    for (int i = 0; i < 3 * n; i++) {
        for (int j = 0; j < 3 * n; j++) {
            for (int k = 0; k < 3 * (n - 1); k++) {
                AtA[i][j] += A[k][i] * A[k][j];
            }
        }
    }

    // 利用高斯消元法求解方程
    vector<double> x(3 * n, 0);
    for (int i = 0; i < 3 * n - 1; i++) {
        // 消元
        for (int j = i + 1; j < 3 * n; j++) {
            double f = AtA[j][i] / AtA[i][i];
            Atb[j] -= f * Atb[i];
            for (int k = i; k < 3 * n; k++) {
                AtA[j][k] -= f * AtA[i][k];
            }
        }
    }
    // 回代
    for (int i = 3 * n - 1; i >= 0; i--) {
        double sum = 0;
        for (int j = i; j < 3 * n; j++) {
            sum += AtA[i][j] * x[j];
        }
        x[i] = (Atb[i] - sum) / AtA[i][i];
    }

    // 将计算结果转换为测站坐标和高程
    vector<Station> result(n);
    for (int i = 0; i < n; i++) {
        result[i].x = x[3 * i];
        result[i].y = x[3 * i + 1];
        result[i].z = x[3 * i + 2] + stations[i].z;
    }

    return result;
}

// 测试代码
int main() {
    // 定义一组测试数据
    vector<Station> stations{{0, 0, 100}, {100, 0, 0}, {0, 100, 0}, {100, 100, 100}};
    vector<Observation> observations{{0, 1, -100, 0, 0}, {0, 2, -100, 0, 90}, {1, 3, -100, 45, 45}, {2, 3, -100, -45, 135}};

    // 调用间接平差函数计算坐标和高程
    vector<Station> result = indirectAdjustment(stations, observations);

    // 输出结果
    for (int i = 0; i < result.size(); i++) {
        cout << "Station " << i << ": (" << result[i].x << ", " << result[i].y << ", " << result[i].z << ")" << endl;
    }

    return 0;
}

需要注意的是，这只是一个简单的实现，没有进行误差分析和精度控制。如果需要更高的精度和可靠性，还需要进行更多的优化和改进。