运输问题表上作业法

[lbp523007444]

// T-Problem_Solver.cpp : Defines the entry point for the console application.
// 表上作业法的源代码

#include "stdafx.h"
#include <fstream>
#include <iomanip>
#include <iostream>
#include <cmath>
using namespace std;
#define a(j) (* (C+(M-1)*N+j)) // 销量数组
#define b(i) (* (C+i*N+N-1)) // 产量数组
#define c(i,j) (* (C+i*N+j)) // 运价数组
#define x(i,j) (* (X+i*(N-1)+j)) // 运量数组
const double BIG_NUM = 1.0E15; // 任意大数
// ( < BIG_NUM: 基本解, >= BIG_NUM : 运量为 0 )
#define s(i,j) (*(S+i*(N-1)+j)) // 检验数数组Sij */
#define u(i) (*(U+i)) // 位势数组 Ui
#define v(i) (*(V+i)) // 位势数组 Vi
#define cpi(k) ((CP+k)->i) // 闭回路点 i 标
#define cpj(k) ((CP+k)->j) // 闭回路点 j 标
#define cpf(k) ((CP+k)->f) // 闭回路点 f 标

/*
f = 0: j++;
f = 1: i--;
f = 2: j--;
f = 3: i++;
*/

void TP(int M,int N,double *C,double *X);

int main()
{
int M, N, i, j;

double* C; // 存储运价, 产量及销量
double* X; // 存储运量分配方案
double z;

ifstream infile;
char fn[80];

double sum;
cout.setf(ios_base::left,ios_base::adjustfield);
cout.setf(ios_base::fixed,ios_base::floatfield);
cout.precision(3);

cout<<"请输入数据文件名: ";
cin>>fn;

infile.open(fn);
if (!infile)
{
cout<<"文件打开失败!\n";
system("pause");
exit(0);
}

infile>>M>>N;

M++;
N++;

X=new double[sizeof(double)*(M-1)*(N-1)];
C=new double[sizeof(double)*M*N];

// 把运价, 供应量和需求量的数据读入到数组 c( i, j )
for(i=0;i<M;++i)
for(j=0;j<N;++j)
{
infile>>z;
c(i,j)=z;
}

infile.close();

cout<<"\n============= 数据文件 ================\n";
for(i=0;i<M;++i)
{
for(j=0;j<N;++j)
cout<<setw(10)<<c(i,j);

cout<<endl;
}

system("pause");

TP(M,N,C,X);

// 输出产销分配方案
cout<<"\n============= 最优解 ===================\n";
sum=0;
for(i=0;i<M-1;++i)
{
for(j=0;j<N-1;++j)
if(x(i,j)>=BIG_NUM)
cout<<setw(10)<<"******";
else
{
cout<<setw(10)<<x(i,j);
sum+=(x(i,j)*c(i,j));
}
cout<<endl;
}

//cout<<"\n\n\tThe min Cost is: %-10.4f\n", sum);
cout<<"\n\n\t最高产量:"<<setw(10)<<sum<<endl; //我们现在是在求max，max=-min


free(X);
free(C);

system("pause");
return 0;
}

// 记录闭回路点结构
struct PATH
{
int i,j,f;
};

void TP(int M,int N,double* C,double* X)
{
double *U, *V, *S;
int MN1,m,n;
struct PATH* CP;
int k,i,j,l,k1,l1,ip;
double Cmin,sum;
int I0,J0,Imin,Jmin;
int fi,fj,fc,f;

MN1=(M-1)+(N-1)-1;
m=M-1;
n=N-1;

S=new double[sizeof(double)*(M-1)*(N-1)];
U=new double[sizeof(double)*M];
V=new double[sizeof(double)*N];
CP=new PATH[sizeof(struct PATH)*(MN1+1)];

// 解初始化 Xij = BIG_NUM
for(i=0;i<m;++i)
for(j=0;j<n;++j)
x(i,j)=BIG_NUM;

// 最小元素法求初始可行解
for ( k = 0; k < MN1; ++k )
{
Cmin = BIG_NUM;

for ( i = 0; i < m; ++i )
{
fi = 0;
for ( l = 0; l < k; ++l )
// 去除已经用过的行
if ( i == cpi( l ) )
{
fi = 1;
break;
}

if ( fi == 1 )
continue;
for ( j = 0; j < n; ++j )
{
fj = 0;
for ( l = 0; l < k; ++l )
// 去除已经用过的列
if ( j == cpj( l ) )
{
fj = 1;
break;
}

if ( fj == 1 )
continue;
if ( Cmin > c( i, j ) )
{
Cmin = c( i, j );
I0 = i;
J0 = j;
}
} // end for j
} // end for i

// 得到了未划去的最小运价所在格的坐标(I0,J0)和最小运价Cmin
if ( k > 0 )
if(Cmin==BIG_NUM && cpi(k-1)==0)
{
for(l1=0;l1<m;l1++)
if(x(l1,cpj(k-1))==BIG_NUM)
x(l1,cpj(k-1))=0;
}
else if(Cmin==BIG_NUM && cpi(k-1)!=0)

for(l1=0;l1<n;l1++)
if(x(cpi(k-1),l1)==BIG_NUM)
x(cpi(k-1),l1)=0;
if ( b( I0 ) < a( J0 ) )
{
cpi( k ) = I0;
cpj( k ) = -1;
x( I0, J0 ) = b( I0 );
a( J0 ) -= b( I0 );
b( I0 ) = 0;
}
else
{
cpi( k ) = -1;
cpj( k ) = J0;
x( I0, J0 ) = a( J0 );
b( I0 ) -= a( J0 );
a( J0 ) = 0;
}
} // end for k 用最小元素法求得了初使可行解
// 输出初始可行解
cout << "\n============= 初始解 ===================\n";
sum = 0;

for ( i = 0; i < M - 1; i++ )
{
for ( j = 0; j < N - 1; j++ )
if ( x( i, j ) >= BIG_NUM )
cout << setw( 10 ) << "******";
else
{
cout << setw( 10 ) << x( i, j );
sum += ( x( i, j ) * c( i, j ) );
}
cout << endl;
}

cout<<"\n\n\t初始产量:"<<setw(10)<<sum<<endl;//我们现在是在求max，max=-min
system( "pause" );

[lbp523007444]

while ( true ) { // 位势置初值 Ui, Vi = BIG_NUM for ( i = 0; i < m; i++ ) u( i ) = BIG_NUM; for ( j = 0; j < n; j++ ) v( j ) = BIG_NUM; // 求位势 l = 0; u( 0 ) = 0; for ( i = 0; i < m; i++ ) for ( j = 0; j < n; j++ ) { if(u(i)>=BIG_NUM && v(j)>=BIG_NUM && x(i,j)<BIG_NUM) { // 记录未求过位势的位置 cpi(l)=i; cpj(l)=j; l++; } else if(x(i,j)<BIG_NUM && u(i)<BIG_NUM) v(j)=c(i,j)-u(i); else if(x(i,j)<BIG_NUM && v(j)<BIG_NUM) u(i)=c(i,j)-v(j); } // 按记录位置求其余位势 if(l>0) while (true) { ip=0; for(k=0;k<l;k++) { i=cpi(k); j=cpj(k); if(u(i)>=BIG_NUM && v(j)>=BIG_NUM && x(i,j)<BIG_NUM) {//记录未求过位势的位置 cpi(ip)=i; cpj(ip)=j; ip++; } else if(x(i,j)<BIG_NUM && u(i)<BIG_NUM) v(j)=c(i,j)-u(i); else if(x(i,j)<BIG_NUM && v(j)<BIG_NUM) u(i)=c(i,j)-v(j); }//end for k if ( ip == 0 ) break; l = ip; } // end for while // 求检验数 for ( i = 0; i < m; i++ ) for(j=0;j<n;j++) { s(i,j)=BIG_NUM; if(x(i,j)>= BIG_NUM) s(i,j)=c(i,j)-u(i)-v(j); } // 求最小检验数 Cmin = BIG_NUM; for(i=0;i<m;i++) for(j=0;j<n;j++) if(Cmin>s(i,j)) { Cmin=s(i,j); I0= i; J0=j; } if ( Cmin >= 0 ) return; // 已经得到最优解,返回主程序 // 此时找到了入基变量 X( I0, J0 ) // 求闭回路 for ( k = 0; k < MN1; k++ ) cpf( k ) = -1; cpi( 0 ) = I0; cpj( 0 ) = J0; k = 0; while ( true ) { f = cpf( k ); // 设置闭回路搜索方向 while ( true ) { i = cpi( k ); j = cpj( k ); fc = 0; f++; cpf( k ) = f; if ( f >= 4 ) break; // 避免反向搜索 if(k>0) { if(f==0 && cpf(k-1)==2) continue; else if(f==1 && cpf(k-1)==3) continue; else if(f==2 && cpf(k-1)==0) continue; else if(f==3 && cpf(k-1)==1) continue;

[lbp523007444]

} if(f==0) { // 沿 j+ 方向搜索 while(true) { j++; if ( j >= n ) { fc = 2; break; } if ( i == I0 && j == J0 ) { fc = 1; break; } if ( s( i, j ) >= BIG_NUM ) { fc = 3; break; } } } // end for j+ else if ( f == 1 ) { // 沿 i- 方向搜索 while ( true ) { i--; if ( i < 0 ) { fc = 2; break; } if ( i == I0 && j == J0 ) { fc = 1; break; } if( s( i, j ) >= BIG_NUM ) { fc = 3; break; } } } // end for i- else if ( f == 2 ) { // 沿 j- 方向搜索 while ( true ) { j--; if( j < 0 ) { fc = 2; break; } if ( i == I0 && j == J0 ) { fc = 1; break; } if( s( i, j ) >= BIG_NUM ) { fc = 3; break; } } } // end for j- else if ( f == 3 ) { // 沿 i+ 方向搜索 while ( true ) { i++; if ( i >= m ) { fc = 2; break; } if ( i == I0 && j == J0 ) { fc = 1; break; } if( s( i, j ) >= BIG_NUM ) { fc = 3; break; } } } // end for i+ if ( fc == 1 || fc == 3 ) break; } // end for while flag 2 if ( fc == 0 ) k--; // 沿此方向搜索失败,退回到前一点 else if ( fc == 1 ) break; // 搜索完成 else if ( fc == 3 ) { // 沿此方向搜索成功,前进一点 k++; cpi( k ) = i; cpj( k ) = j; cpf( k ) = -1; } } // end while // 去除闭回路中的非转折点 l = 0; while ( l < k - 1 ) { i = l + 1; while ( i <= k ) { if( cpf( l ) == cpf( i ) ) i++; else break; } if ( i > ( l + 1 ) ) { j = l + 1; k1 = k - ( i - j ); // 如果某些点前进方向相同,则去除中间各点 while( i <= k ) { cpi( j ) = cpi( i ); cpj( j ) = cpj( i ); cpf( j ) = cpf( i ); i++; j++; } l += 2; k = k1; } else l++; } // end for while l < k - 1 // 查找闭回路上基本解的最小值 Cmin = x( cpi( 1 ), cpj( 1 ) ); Imin = cpi( 1 ); Jmin = cpj( 1 ); for ( i = 3; i <= k; i += 2 ) if( Cmin > x( cpi( i ), cpj( i ) ) ) { Cmin = x( cpi( i ),cpj( i ) ); Imin = cpi( i ); Jmin = cpj( i ); } // 换入基变量 x( I0, J0 ) = Cmin; for( i = 1; i <= k; i += 2 ) { x( cpi( i ), cpj( i ) ) -= Cmin; cout<<"hao"<<x( cpi( i ), cpj( i ) ); if ( ( i + 1 ) <= k ) x(cpi(i+1),cpj(i+1))+=Cmin; } x( Imin, Jmin ) = BIG_NUM; } delete [] CP; delete [] V; delete [] U; delete [] S; } //以上是我下载的源代码，下面的是程序读入的数据 9 3 3 M 2 30 M 1 M 10 3 M 2 12 3 M 2 3 3 M 2 9 M 1 M 14 3 M 2 20 M 1 M 18 M M M 4 40 40 40 //为什么算出的结果与手工算的结果相差甚大？

[hugett]