单纯形法原理，解决问题领域，C#及C++代码

单纯形法是一种线性规划算法，用于在一组约束条件下最小化或最大化一个线性函数。该算法通过不断迭代计算，逐步接近最优解。

单纯形法可以应用于许多领域，如生产规划、资源分配、运输问题等。

以下是C#和C++中的单纯形法代码示例：

C#代码：

using System;

class Simplex {
    static void Main() {
        // 定义线性规划问题
        double[][] A = { // 约束条件系数矩阵
            new double[] {  1,  1,  0,  1,  0,  0 },
            new double[] {  2,  1,  1,  0,  1,  0 },
            new double[] {  1,  2, -1,  0,  0,  1 }
        };
        double[] b = { 4, 7, 3 }; // 约束条件右侧常数向量
        double[] c = { 3, 2, 1, 0, 0, 0 }; // 目标函数系数向量

        // 使用单纯形法求解线性规划问题
        SimplexSolver solver = new SimplexSolver();
        bool success = solver.Solve(A, b, c);

        // 输出结果
        if (success) {
            Console.WriteLine("最优解: " + solver.OptimumValue);
            Console.WriteLine("最优解向量: " + string.Join(", ", solver.OptimumSolution));
        } else {
            Console.WriteLine("无可行解");
        }
    }
}

class SimplexSolver {
    private const double EPSILON = 1e-10; // 误差范围
    private int m, n; // 约束条件数和变量数
    private int[] basis; // 基变量下标
    private double[][] A; // 约束条件系数矩阵
    private double[] b; // 约束条件右侧常数向量
    private double[] c; // 目标函数系数向量
    private double Optimum; // 最优解
    private double[] OptimumSolution; // 最优解向量

    public bool Solve(double[][] A, double[] b, double[] c) {
        this.A = A;
        this.b = b;
        this.c = c;
        m = A.Length;
        n = A[0].Length;
        basis = new int[m];

        // 初始化基变量下标
        for (int i = 0; i < m; i++) {
            basis[i] = n + i;
        }

        // 标准化线性规划问题
        var s = new double[m][];
        for (int i = 0; i < m; i++) {
            s[i] = new double[n + m + 1];
            for (int j = 0; j < n; j++) {
                s[i][j] = A[i][j];
            }
            s[i][n + i] = 1;
            s[i][n + m] = b[i];
        }
        n += m;
        this.A = s;

        // 迭代计算
        while (true) {
            int q = 0;
            for (int i = 0; i < n; i++) {
                if (c[i] > c[q]) {
                    q = i;
                }
            }
            if (c[q] < EPSILON) {
                break;
            }
            int p = -1;
            double min = double.MaxValue;
            for (int i = 0; i < m; i++) {
                if (A[i][q] > EPSILON && b[i] / A[i][q] < min) {
                    min = b[i] / A[i][q];
                    p = i;
                }
            }
            if (p == -1) {
                return false;
            }
            Pivot(p, q);
        }

        // 计算最优解及最优解向量
        Optimum = -A[m][n];
        OptimumSolution = new double[n];
        for (int i = 0; i < m; i++) {
            if (basis[i] < n) {
                OptimumSolution[basis[i]] = A[i][n];
            }
        }

        return true;
    }

    private void Pivot(int p, int q) {
        for (int i = 0; i < m; i++) {
            if (i != p) {
                double alpha = A[i][q] / A[p][q];
                for (int j = 0; j <= n; j++) {
                    A[i][j] -= alpha * A[p][j];
                }
            }
        }
        for (int j = 0; j <= n; j++) {
            if (j != q) {
                A[p][j] /= A[p][q];
            }
        }
        A[p][q] = 1;
        basis[p] = q;
    }

    public double OptimumValue {
        get { return Optimum; }
    }

    public double[] OptimumSolution {
        get { return OptimumSolution; }
    }
}

C++代码：

#include <iostream>
#include <cstring>
#include <cmath>

using namespace std;

const double EPS = 1e-10;

class Simplex {
public:
    Simplex(int m, int n, double** A, double* b, double* c) {
        this->m = m;
        this->n = n;
        this->A = A;
        this->b = b;
        this->c = c;
        basis = new int[m];
        for (int i = 0; i < m; i++) {
            basis[i] = n + i;
        }
    }

    bool Solve() {
        while (true) {
            int q = 0;
            for (int i = 0; i < n; i++) {
                if (c[i] > c[q]) {
                    q = i;
                }
            }
            if (c[q] < EPS) {
                break;
            }
            int p = -1;
            double min = 1e9;
            for (int i = 0; i < m; i++) {
                if (A[i][q] > EPS && b[i] / A[i][q] < min) {
                    min = b[i] / A[i][q];
                    p = i;
                }
            }
            if (p == -1) {
                return false;
            }
            Pivot(p, q);
        }
        while (true) {
            int p = -1;
            double min = 0;
            for (int i = 0; i < m; i++) {
                if (A[i][n] < min) {
                    min = A[i][n];
                    p = i;
                }
            }
            if (p == -1) {
                break;
            }
            int q = 0;
            for (int i = 0; i < n; i++) {
                if (A[p][i] < -EPS && c[i] / A[p][i] < c[q] / A[p][q]) {
                    q = i;
                }
            }
            if (A[p][q] > -EPS) {
                return false;
            }
            Pivot(p, q);
        }
        return true;
    }

    double OptimumValue() {
        return -A[m][n];
    }

    double* OptimumSolution() {
        double* x = new double[n];
        memset(x, 0, sizeof(double) * n);
        for (int i = 0; i < m; i++) {
            if (basis[i] < n) {
                x[basis[i]] = A[i][n];
            }
        }
        return x;
    }

private:
    void Pivot(int p, int q) {
        for (int i = 0; i < m; i++) {
            if (i != p) {
                double alpha = A[i][q] / A[p][q];
                for (int j = 0; j <= n; j++) {
                    A[i][j] -= alpha * A[p][j];
                }
            }
        }
        for (int j = 0; j <= n; j++) {
            if (j != q) {
                A[p][j] /= A[p][q];
            }
        }
        A[p][q] = 1;
        basis[p] = q;
    }

    int m, n;
    double** A;
    double* b;
    double* c;
    int* basis;
};

int main() {
    double A[3][6] = {
        { 1, 1, 0, 1, 0, 0 },
        { 2, 1, 1, 0, 1, 0 },
        { 1, 2, -1, 0, 0, 1 }
    };
    double b[3] = { 4, 7, 3 };
    double c[6] = { 3, 2, 1, 0, 0, 0 };
    Simplex simplex(3, 6, (double**)A, b, c);
    if (simplex.Solve()) {
        cout << "最优解: " << simplex.OptimumValue() << endl;
        cout << "最优解向量: ";
        double* x = simplex.OptimumSolution();
        for (int i = 0; i < 6; i++) {
            cout << x[i] << " ";
        }
        cout << endl;
    } else {
        cout << "无可行解" << endl;
    }
    return 0;
}