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c++實現單純形法現行規劃問題的求解(推薦)

在本程式中預設該現行規劃問題有最優解

針對此問題:

c++實現單純形法現行規劃問題的求解(推薦)

#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;
}

//此程式已經化為標準型的線性規劃問題中,且預設有最優解
int main(int argc,char* argv[])
{
  //資料輸入部分
  int m,n;
  cout << "請輸入變數個數:";
  cin >> m;
  cout << "請輸入不等式個數:";
  cin >> n;
  float **matrix = new float*[n + 1];   //係數矩陣
  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];  //基變數係數
  int *XB = new int[n + 1];  //用來標註基變數x的下標
  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 << "請輸入目標函式係數(用空格間開):" << endl;
  for (int i = 1; i <= m; i++) {
    cin >> cj[i];
  }
  cout << "請輸入各不等式的係數和常量(用空格間開):" << 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 << "請輸入目標函式中基變數下標:" << endl;
  for (int i = 1; i <= n; i++) {
    cin >> XB[i];
    cB[i] = cj[XB[i]];
    //常量
    b[i] = matrix[i][m + 1];
  }

  //計算檢驗數
  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) {
    //尋找入基變數
  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];
    }
  }
  //計算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值為負數
    }
  }
  //尋找出基變數
  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;
    }
  }
  //入基出基變換,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;
    }
  }
  //出基計算 
  //化1
  //cout << endl << "此處為化1的結果------" << 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 << "此處為化0的結果------" << 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 << "最優解為: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 << "最優值為:MzxZ = " << MaxZ;
  return 0;
}

程式執行結果:

c++實現單純形法現行規劃問題的求解(推薦)

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