c++实现单纯形法现行规划问题的求解(推荐)

发布时间:2020-09-08 17:18:45 作者:含光Aries
来源:脚本之家 阅读:174

在本程序中默认该现行规划问题有最优解

针对此问题:

c++实现单纯形法现行规划问题的求解(推荐)

#include<iostream>
using namespace std;

int check(float *sigema, int m) {
  for (int i = 1; i <= m ; i++) {
    if (sigema[i] > 0) {
      return 0;
    }
  }
  return 1;
}

//此程序已经化为标准型的线性规划问题中,且默认有最优解
int main(int argc, char* argv[])
{
  //数据输入部分
  int m, n;
  cout << "请输入变量个数:";
  cin >> m;
  cout << "请输入不等式个数:";
  cin >> n;
  float **matrix = new float*[n + 1];   //系数矩阵
  for (int i = 1; i <= n; i++) {
    matrix[i] = new float[m + 2];
  }
  float *cj = new float[m + 1];
  float *cB = new float[n + 1];  //基变量系数
  int *XB = new int[n + 1];  //用来标注基变量x的下标
  float *b = new float[n + 1];
  float *sigema = new float[n + 1];
  float *sita = new float[n + 1];
  //初始化
  for (int i = 0; i <= m; i++) {
    cj[i] = 0;
  }
  for (int i = 0; i <= n; i++) {
    cB[i] = 0;
    XB[i] = 0;
    b[i] = 0;
    sigema[i] = 0;
    sita[i] = 0;
  }
  cout << "请输入目标函数系数(用空格间开):" << endl;
  for (int i = 1; i <= m; i++) {
    cin >> cj[i];
  }
  cout << "请输入各不等式的系数和常量(用空格间开):" << endl;
  for (int i = 1; i <= n; i++) {
    cout << "不等式" << i << ": ";
    for (int j = 1; j <= m + 1; j++) {
      cin >> matrix[i][j];
    }
  }
  cout << "请输入目标函数中基变量下标:" << endl;
  for (int i = 1; i <= n; i++) {
    cin >> XB[i];
    cB[i] = cj[XB[i]];
    //常量
    b[i] = matrix[i][m + 1];
  }

  //计算检验数
  for (int i = 1; i <= m; i++) {
    sigema[i] = cj[i];
    for (int j = 1; j <= n; j++) {
      sigema[i] -= cB[j] * matrix[j][i];
    }
  }

  while (check(sigema, m) == 0) {
    //寻找入基变量
  float maxn = sigema[1];
  int sigema_xindex = 0;
  float sigema_xcoefficient = 0;
  for (int i = 1; i <= m; i++) {
    if (maxn <= sigema[i]) {
      maxn = sigema[i];
      sigema_xindex = i;
      sigema_xcoefficient = cj[i];
    }
  }
  //计算sita
  for (int i = 1; i <= n; i++) {
    if (matrix[i][sigema_xindex] > 0) {
      sita[i] = b[i] / matrix[i][sigema_xindex];
    }
    else {
      sita[i] = 9999; //表示sita值为负数
    }
  }
  //寻找出基变量
  float minn = sita[1];
  int sita_xindex = 0;
  for (int i = 1; i <= n; i++) {
    if (minn >= sita[i] && sita[i] > 0) {
      minn = sita[i];
      sita_xindex = i;
    }
  }
  //入基出基变换,先入基再出基
  //入基操作
  for (int i = 1; i <= n; i++) {
    if (i == sita_xindex) {
      XB[i] = sigema_xindex;
      cB[i] = sigema_xcoefficient;
      break;
    }
  }
  //出基计算 
  //化1
  //cout << endl << "此处为化1的结果------" << endl;
  float mul1 = matrix[sita_xindex][sigema_xindex];
  for (int i = 1; i <= m; i++) {
    matrix[sita_xindex][i] /= mul1;
  }
  b[sita_xindex] /= mul1;
  //化0
  //cout << endl << "此处为化0的结果------" << endl;
  for (int i = 1; i <= n; i++) {
    if (i == sita_xindex) {
      continue;
    }
    float mul2 = matrix[i][sigema_xindex] / matrix[sita_xindex][sigema_xindex];
    for (int j = 1; j <= m; j++) {
      matrix[i][j] -= (matrix[sita_xindex][j] * mul2);
    }
    b[i] -= (b[sita_xindex] * mul2);
  }
  for (int i = 1; i <= n; i++) {
    if (i == sita_xindex) {
      continue;
    }
  }
  for (int i = 1; i <= m; i++) {
    sigema[i] = cj[i];
    for (int j = 1; j <= n; j++) {
      sigema[i] -= cB[j] * matrix[j][i];
    }
  }
  }
  float MaxZ = 0;
  float *result = new float[m + 1];
  for (int i = 0; i <= m; i++) {
    result[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    result[XB[i]] = b[i];
  }
  cout << "最优解为:X = (";
  for (int i = 1; i < m; i++) {
    cout << result[i] << ",";
  }
  cout << result[m] << ")" << endl;
  for (int i = 1; i <= m; i++) {
    MaxZ += result[i] * cj[i];
  }
  cout << "最优值为:MzxZ = " << MaxZ;
  return 0;
}

程序运行结果:

c++实现单纯形法现行规划问题的求解(推荐)

总结

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