我无法找到单纯形法的实现。我有一组点并希望最小化距离所以我只需要方法simplex 在发布此问题之前我已经谷歌了，并且找不到任何可以使用的内容
答案 0 :(得分：30)
/*
What: Simplex in C
AUTHOR: GPL(C) moshahmed/at/gmail.
What: Solves LP Problem with Simplex:
{ maximize cx : Ax <= b, x >= 0 }.
Input: { m, n, Mat[m x n] }, where:
b = mat[1..m,0] .. column 0 is b >= 0, so x=0 is a basic feasible solution.
c = mat[0,1..n] .. row 0 is z to maximize, note c is negated in input.
A = mat[1..m,1..n] .. constraints.
x = [x1..xm] are the named variables in the problem.
Slack variables are in columns [m+1..m+n]
USAGE:
1. Problem can be specified before main function in source code:
c:\> vim mosplex.c
Tableau tab = { m, n, { // tableau size, row x columns.
{ 0 , -c1, -c2, }, // Max: z = c1 x1 + c2 x2,
{ b1 , a11, a12, }, // b1 >= a11 x1 + a12 x2
{ b2 , a21, a22, }, // b2 >= a21 x1 + a22 x2
}
};
c:\> cl /W4 mosplex.c ... compile this file.
c:\> mosplex.exe problem.txt > solution.txt
2. OR Read the problem data from a file:
$ cat problem.txt
m n
0 -c1 -c2
b1 a11 a12
b2 a21 a11
$ gcc -Wall -g mosplex.c -o mosplex
$ mosplex problem.txt > solution.txt
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <assert.h>
#define M 20
#define N 20
static const double epsilon = 1.0e-8;
int equal(double a, double b) { return fabs(a-b) < epsilon; }
typedef struct {
int m, n; // m=rows, n=columns, mat[m x n]
double mat[M][N];
} Tableau;
void nl(int k){ int j; for(j=0;j<k;j++) putchar('-'); putchar('\n'); }
void print_tableau(Tableau *tab, const char* mes) {
static int counter=0;
int i, j;
printf("\n%d. Tableau %s:\n", ++counter, mes);
nl(70);
printf("%-6s%5s", "col:", "b[i]");
for(j=1;j<tab->n; j++) { printf(" x%d,", j); } printf("\n");
for(i=0;i<tab->m; i++) {
if (i==0) printf("max:"); else
printf("b%d: ", i);
for(j=0;j<tab->n; j++) {
if (equal((int)tab->mat[i][j], tab->mat[i][j]))
printf(" %6d", (int)tab->mat[i][j]);
else
printf(" %6.2lf", tab->mat[i][j]);
}
printf("\n");
}
nl(70);
}
/* Example input file for read_tableau:
4 5
0 -0.5 -3 -1 -4
40 1 1 1 1
10 -2 -1 1 1
10 0 1 0 -1
*/
void read_tableau(Tableau *tab, const char * filename) {
int err, i, j;
FILE * fp;
fp = fopen(filename, "r" );
if( !fp ) {
printf("Cannot read %s\n", filename); exit(1);
}
memset(tab, 0, sizeof(*tab));
err = fscanf(fp, "%d %d", &tab->m, &tab->n);
if (err == 0 || err == EOF) {
printf("Cannot read m or n\n"); exit(1);
}
for(i=0;i<tab->m; i++) {
for(j=0;j<tab->n; j++) {
err = fscanf(fp, "%lf", &tab->mat[i][j]);
if (err == 0 || err == EOF) {
printf("Cannot read A[%d][%d]\n", i, j); exit(1);
}
}
}
printf("Read tableau [%d rows x %d columns] from file '%s'.\n",
tab->m, tab->n, filename);
fclose(fp);
}
void pivot_on(Tableau *tab, int row, int col) {
int i, j;
double pivot;
pivot = tab->mat[row][col];
assert(pivot>0);
for(j=0;j<tab->n;j++)
tab->mat[row][j] /= pivot;
assert( equal(tab->mat[row][col], 1. ));
for(i=0; i<tab->m; i++) { // foreach remaining row i do
double multiplier = tab->mat[i][col];
if(i==row) continue;
for(j=0; j<tab->n; j++) { // r[i] = r[i] - z * r[row];
tab->mat[i][j] -= multiplier * tab->mat[row][j];
}
}
}
// Find pivot_col = most negative column in mat[0][1..n]
int find_pivot_column(Tableau *tab) {
int j, pivot_col = 1;
double lowest = tab->mat[0][pivot_col];
for(j=1; j<tab->n; j++) {
if (tab->mat[0][j] < lowest) {
lowest = tab->mat[0][j];
pivot_col = j;
}
}
printf("Most negative column in row[0] is col %d = %g.\n", pivot_col, lowest);
if( lowest >= 0 ) {
return -1; // All positive columns in row[0], this is optimal.
}
return pivot_col;
}
// Find the pivot_row, with smallest positive ratio = col[0] / col[pivot]
int find_pivot_row(Tableau *tab, int pivot_col) {
int i, pivot_row = 0;
double min_ratio = -1;
printf("Ratios A[row_i,0]/A[row_i,%d] = [",pivot_col);
for(i=1;i<tab->m;i++){
double ratio = tab->mat[i][0] / tab->mat[i][pivot_col];
printf("%3.2lf, ", ratio);
if ( (ratio > 0 && ratio < min_ratio ) || min_ratio < 0 ) {
min_ratio = ratio;
pivot_row = i;
}
}
printf("].\n");
if (min_ratio == -1)
return -1; // Unbounded.
printf("Found pivot A[%d,%d], min positive ratio=%g in row=%d.\n",
pivot_row, pivot_col, min_ratio, pivot_row);
return pivot_row;
}
void add_slack_variables(Tableau *tab) {
int i, j;
for(i=1; i<tab->m; i++) {
for(j=1; j<tab->m; j++)
tab->mat[i][j + tab->n -1] = (i==j);
}
tab->n += tab->m -1;
}
void check_b_positive(Tableau *tab) {
int i;
for(i=1; i<tab->m; i++)
assert(tab->mat[i][0] >= 0);
}
// Given a column of identity matrix, find the row containing 1.
// return -1, if the column as not from an identity matrix.
int find_basis_variable(Tableau *tab, int col) {
int i, xi=-1;
for(i=1; i < tab->m; i++) {
if (equal( tab->mat[i][col],1) ) {
if (xi == -1)
xi=i; // found first '1', save this row number.
else
return -1; // found second '1', not an identity matrix.
} else if (!equal( tab->mat[i][col],0) ) {
return -1; // not an identity matrix column.
}
}
return xi;
}
void print_optimal_vector(Tableau *tab, char *message) {
int j, xi;
printf("%s at ", message);
for(j=1;j<tab->n;j++) { // for each column.
xi = find_basis_variable(tab, j);
if (xi != -1)
printf("x%d=%3.2lf, ", j, tab->mat[xi][0] );
else
printf("x%d=0, ", j);
}
printf("\n");
}
void simplex(Tableau *tab) {
int loop=0;
add_slack_variables(tab);
check_b_positive(tab);
print_tableau(tab,"Padded with slack variables");
while( ++loop ) {
int pivot_col, pivot_row;
pivot_col = find_pivot_column(tab);
if( pivot_col < 0 ) {
printf("Found optimal value=A[0,0]=%3.2lf (no negatives in row 0).\n",
tab->mat[0][0]);
print_optimal_vector(tab, "Optimal vector");
break;
}
printf("Entering variable x%d to be made basic, so pivot_col=%d.\n",
pivot_col, pivot_col);
pivot_row = find_pivot_row(tab, pivot_col);
if (pivot_row < 0) {
printf("unbounded (no pivot_row).\n");
break;
}
printf("Leaving variable x%d, so pivot_row=%d\n", pivot_row, pivot_row);
pivot_on(tab, pivot_row, pivot_col);
print_tableau(tab,"After pivoting");
print_optimal_vector(tab, "Basic feasible solution");
if(loop > 20) {
printf("Too many iterations > %d.\n", loop);
break;
}
}
}
Tableau tab = { 4, 5, { // Size of tableau [4 rows x 5 columns ]
{ 0.0 , -0.5 , -3.0 ,-1.0 , -4.0, }, // Max: z = 0.5*x + 3*y + z + 4*w,
{ 40.0 , 1.0 , 1.0 , 1.0 , 1.0, }, // x + y + z + w <= 40 .. b1
{ 10.0 , -2.0 , -1.0 , 1.0 , 1.0, }, // -2x - y + z + w <= 10 .. b2
{ 10.0 , 0.0 , 1.0 , 0.0 , -1.0, }, // y - w <= 10 .. b3
}
};
int main(int argc, char *argv[]){
if (argc > 1) { // usage: cmd datafile
read_tableau(&tab, argv[1]);
}
print_tableau(&tab,"Initial");
simplex(&tab);
return 0;
}