对偶单纯型法解线性规划问题

本程序实现的是单纯型法解线性规划问题，在检验向量e<=0的情况下，迭代得到最优解。

首先，要求你在data.txt文件中输入数据，输入方式，例：

 

 在data.txt中输入：

3
1 1 1
2 3
 3 1 1 >=  1
-1 4 1 >=  2
 

便可出现执行结果!

实现程序如下:

#include<iostream>
#include<fstream>
#include<string>
#include<cstdlib>
#define INF -100000
#define MAX 100000
using namespace std;
int N1, N2;
//约束条件系数
double **a;
double *b; // 向量b
double *e;//检验数向量
double RHS = 0;//目标函数值
int *x; // 存放基本可行解
int m;  // 添加的辅助变量的个数
int JudgeSym(string symbol)
{
    if(symbol==">=")
      return -1;
    if(symbol=="<=")
      return 1;
    if(symbol=="=")
      return 0;
}
// 找出初值解
void Begin()
{
     //进行列搜索， 得到初始解
    int index = 0;
    for(int j=1; j<N2+m; ++j){
      int cnt = 0;
     for(int i=1; i<N1; ++i)
      if(a[i][j]){
         ++cnt;
         index = i;
       }
      if(cnt==1&&x[index]==0){
         if(a[index][j] != 1){
             for(int k=1; k<N2; ++k)
               if(k!=j)
                 a[index][k] /= a[index][j];
             b[index] /= a[index][j];
             a[index][j] = 1;
         }
        x[index] = j;
      }
    }
    cout<<"得到对偶问题的初始基本可行解为:"<<endl;
    for(int i=1; i<N1; ++i)
        cout<<"x("<<x[i]<<")"<<" = "<<b[i]/a[i][x[i]]<<endl;
}
void GetFirstForm()
{
     double tmp = 0;
     for(int j=1; j<N1; ++j)
         if(e[x[j]]!=0){
            tmp = e[x[j]];
         for(int i=1; i<N2; ++i){
             /*May Have  Error */
             e[i] -=   tmp*a[j][i];
            }
         RHS -= tmp*b[j];
     }
}
int CheckRvec()
{
    double min = MAX;
    int index = 0;
    for(int i=1; i<N1; ++i)
      if(b[i]<0&&b[i]<min){
       min =b[i];
       index = i;
      }
    if(index!=0)
      return index;
    cout<<"迭代得到一个最优解:RHS = "<<RHS<<endl;
    for(int i=1; i<N1; ++i)
      cout<<"x("<<x[i]<<")"<<" = "<<b[i]/a[i][x[i]]<<endl;
    return -1;
}
int MinEj(int Xiout)
{
    double min = MAX;
    int index = -1;
    for(int j=1; j<N2+m; ++j)
      if(a[Xiout][j]<0&&e[j]/a[Xiout][j]<min){
        index = j;
        min = e[j]/a[Xiout][j];
      }
    if(index == -1)
     cout<<"a(r,j)>=0;无最优解!"<<endl;
    return index;
}
void InputandInit()
{
    void Standardlize(int,int);
    ifstream infile("data.txt");
    infile>>N2;
    N2 += 1;
    //存储没有添加辅助变量的
    double *tpe = new double[N2];
    for(int i=1; i<N2; ++i){
     infile>>tpe[i];
     tpe[i] = -tpe[i];
     }/*修改，根据BeginOne()的返回值，判断是否进行e的再次录入*/
    infile>>N1>>N2;
    m = N1; /*最多添加的辅助变量的个数*/
    N1 += 1;
    N2 += 1;
    //正式进行目标函数的录入
    e = new double[N2+m];
    memset(e,0,sizeof(double)*(N2+m));
    memcpy(e,tpe,sizeof(double)*N2);
    delete tpe;/**/
    //为 系数矩阵申请内存
    a = new double*[N1];
    for(int i=1; i<N1; ++i){
      a[i] = new double[N2+m]; /*a : N1/N2+N1 */
      memset(a[i],0,sizeof(double)*(N2+m));
    }
    //为 右端向量申请内存
    b = new double[N1];
    memset(b,0,sizeof(double)*N1);

    m = 0; /*申请内存结束后,重新置m为零,使m记录实际添加的人工变量的个数*/

    //输入a系数矩阵
    for(int i=1; i<N1; ++i){
     for(int j=1; j<N2; ++j){
       infile>>a[i][j];
     }
     string Symbol;
     infile>>Symbol;
     infile>>b[i];
     int numSym = JudgeSym(Symbol);
     Standardlize(numSym,i);
     }
}
void Standardlize(int numSym, int line)
{
     int row = N1 -1;
     int col = N2 -1;
     static int cnt = 1;
     switch(numSym){
      case -1:
        for(int j=1; j<N2; ++j)
          a[line][j] = -a[line][j];
        b[line] = -b[line];
        a[line][col+cnt] = 1;/*添加辅助变量*/
        ++cnt;
        ++m;  /*辅助变量加1*/
        break;
      case 1:
        a[line][col+cnt] = 1;
         ++cnt;
         ++m; /*辅助变量加1*/
        break;
      default: return;
     }
     return;
}
void print()
{
     static int cnt = 0;
     cout<<left<<showpos;
     cout.width(5);
     cout.precision(3);
     cout<<cnt<<"-Z  ";
    for(int i=1; i<N2+m; ++i)
      cout<<e[i]<<" ";
      cout<<RHS<<endl;
    for(int i=1; i<N1; ++i){
      cout<<"x("<<x[i]<<") ";
      for(int j=1; j<N2+m; ++j)
        cout<<a[i][j]<<" ";
      cout<<b[i]<<endl;
    }
    cout<<"***********************"<<endl;
    ++cnt;
}
int main()
{
    InputandInit();
    /* Until then; data input finished */
    x = new int[N1];
    memset(x,0,sizeof(int)*N1);
    Begin();
    GetFirstForm();
    //打印第一张单纯行表
    cout<<"The First Form Of Simplex Method:"<<endl;
    print();
    /*循环得出最优解*/
    int Xiout = 0, ej = 0;           /*找到最小的e(j)/a(Xiout)(j)*/
    while((Xiout=CheckRvec())!=-1 &&(ej = MinEj(Xiout))!=-1){
        x[Xiout] = ej;
        for(int i=1; i<N2+m; ++i)
          if(ej!=i){
           a[Xiout][i] = a[Xiout][i]/a[Xiout][ej];
           }
        /*Error will be found*/
        b[Xiout]/=a[Xiout][ej];
        a[Xiout][ej] = 1;
        //化检验数向量
        double tmp = e[ej];
        for(int i=1; i<N2+m; ++i)
           e[i] -= tmp*a[Xiout][i];
        RHS -= tmp*b[Xiout];
        /*化右端向量*/
        for(int i=1; i<N1; ++i){
          if(i!=Xiout){
           tmp = a[i][ej];
          for(int j=1; j<N2+m; ++j)
            a[i][j] -= tmp*a[Xiout][j];
          b[i] -= tmp*b[Xiout];
          }
        }
        print();
    }
    system("pause");
    return 0;
}