单纯形法C++代码

//X[k] enters and X[l] leaves
XB[l]=k;
CB[l]=C[k];
//row calculation
float bl=0;
float alk=0;
float aik=0;
B[l]/=A[l][k];
bm am1 am2 am3...amn
*/
#include<iostream>
#include <iomanip>
using namespace std;
#define Max 100
#define Epsilon 0பைடு நூலகம்0000001
/*----------------1-variables statement begins------------------*/
}
cout<<" else X= 0.00."<<endl<<endl;
cout.unsetf(ios::showpos);
}
else if (a==2)//unbounded solution
cout<<"Unbounded solution!"<<endl<<endl;
float z=0;//object value
bool flag=false;//flag whether to stop pivoting
/*----------------1-variables statement ends------------------*/
/*----------------2-function of input begins------------------*/
int count=0;//number of linear problems
int n=0;//number of variables
int N=0;//number of slack variables
int m=0;//number of constraints
int XB[Max]={0};//basic variable
}
//}
}
cout<<endl;
}
cout<<"----------------------------------------------------------"<<endl;
//print Z | X-XB
cout<<"Z ="<<fixed<<setprecision(2)<<showpos<<z<<" ";
for ( i=0; i<m; i++)
{
cin>>B[i];
for (int j=0; j<N; j++)
cin>>A[i][j];
A[i][N+i]=1;
}
//obtain slack variable
for ( i=0; i<m; i++)
XB[i]=N+i;
void print()
{
z=0;
for( int i=0; i<m; i++)
z+=CB[i]*B[i];
//print XB[m] | B[m] A[m][n]
for( i=0; i<m; i++)
{
cout.unsetf(ios::showpos);
cout<<"X"<<XB[i]+1<<"= "<<fixed<<setprecision(2)<<B[i]<<" ";
for(i=0; i<n; i++)
{
if(Delta[i]!=0)
{
cout<<fixed<<setprecision(2)<<showpos<<Delta[i]<<"X";
cout.unsetf(ios::showpos);
cout<<(i+1)<<" ";
cout.unsetf(ios::showpos);
}
/*------------------3-function of output ends--------------------*/
/*------------------4-function of dictionary printing begins--------------------*/
cout<<"The optimal solution: "<<fixed<<setprecision(2)<<showpos<<z<<endl<<endl;
for (i=0; i<m; i++)
{
cout.unsetf(ios::showpos);
cout<<"X"<<XB[i]+1<<"= "<<fixed<<setprecision(2)<<showpos<<B[i]<<", ";
void output(int a)
{
flag=false;//flag to stop next pivoting
if (a==1)//output the optimal solution
{
z=0;
for(int i=0; i<m; i++)
z+=CB[i]*B[i];
A[i][j]=0;
}
//input m & n
if(!(cin>>N>>m))
return false;
n=N+m;
//input c[i]
for ( i=0; i<N; i++)
cin>>C[i];
//input b[i] & a[i][j]
for(int j=0; j<n; j++)
{
bool temp3=true;
//if(A[i][j]!=0)
//{
for(int k=0; k<m; k++)
{
if (j==XB[k])
{
temp3=false;
//judge according to the dictionary whether the solution is the optimal
if(num==n)
{
output(1);
return;
}
//find the entering variable X[k]
{
output(3);
return;
}
temp=temp+CB[j]*A[j][i];
}
Delta[i]=C[i]-temp;
if (Delta[i]<=0)
num++;
}
//print dictionary
print();
output(2);
return;
}
cout<<"X"<<k+1<<" enters. ";
//find the leaving variable X[l]
int l=0;
int Xl=0;
int temp2=0;
float min=Max;
for ( i=0; i<m; i++)
void simplex()
{
int num=0;
//calculate checking values-1
for (int i=0; i<n; i++)
{
float temp=0;
for (int j=0; j<m; j++)
{
if (B[j]<Epsilon)//B[i]<0, infeasible initial dictionary
break;
}
}
if(temp3)
{
cout<<fixed<<setprecision(2)<<showpos<<-A[i][j]<<"X";
cout.unsetf(ios::showpos);
cout<<(j+1)<<" ";
float al[Max]={0};//temp storing vector
float Delta[Max]={0};//checking values-1:for entering variable decision