C Implementation for Matrix Method using Transportation Problem SCM

Scm_matrix.c

/**
 * @main-author Tomrock D'souza, St. Francis Institute Of Technology, University of Mumbai, 2017
 * Email: [email protected] 
 * No reproduction in whole or part without maintaining this notice
 * and imposing this condition on any subsequent users.
 * Example: https://postimg.org/image/y55004t9z/
 */
#include<stdio.h>
#include<conio.h>


/*
 * storMat is te matrix that stores values of the numerical.
 * numFact is the Max Factories or Importers
 * numSupp is the Max Exporters
 */
int storMat[30][30], i, j, numSupp, numFact;


/*
 * checker function checks if either
 * all importer Demands are satisfied or
 * all Exporters have no supply after each iterraion
 */
int checker() {
  int a=0,b=0;
  for (i = 0; i <= numSupp; i++) {
    if (storMat[i][numFact] != 0) {
      a = 1;
	  break;
    }
  }
  for (i = 0; i <= numFact; i++) {
    if (storMat[numSupp][i] != 0) {
      b = 1;
	  break;
    }
  }
  return a+b;
}


/*
 * printr function prints the Numerical's current status
 */
void printr() {

for (i = 0; i < numSupp; i++) {
for (j = 0; j < numFact; j++) {
      printf("%d\t", storMat[i][j]);
    }
	printf("| %d", storMat[i][j]);
    printf("\n");
	
  }
    for(j=0;j<=8*numFact;j++){printf("-");}
	printf("\n");
	for (j = 0; j < numFact; j++) {
      printf("%d\t", storMat[numSupp][j]);
    }
	printf("\n");
}


/*
 * Checks the next avialable minimum in the list and performs and iteration returning it's amount 
 */
long checkMin() {
  int a=9999, b,k1,k2;
  long g;
    for (i = 0; i < numSupp; i++) {
      for (j = 0; j < numFact; j++) {
		if ((storMat[i][j] < a) &&(storMat[numSupp][j] > 0)&&(storMat[i][numFact] > 0)) {
        k1 = i;
        k2 = j;
		a = storMat[k1][k2];
      }
    }
  }
/*
 * At this stage the least cost Exporter is found for this particular iteration
 * Further the deduction operations will be performed
 */
  if (storMat[k1][numFact] < storMat[numSupp][k2]) {
    storMat[numSupp][k2] -= storMat[k1][numFact];
    b = storMat[k1][numFact];
    storMat[k1][numFact] = 0;
  } else if (storMat[k1][numFact] > storMat[numSupp][k2]) {
    storMat[k1][numFact] -= storMat[numSupp][k2];
    b = storMat[numSupp][k2];
    storMat[numSupp][k2] = 0;;
  } else {
    b = storMat[k1][numFact];
    storMat[k1][numFact] = 0;
    storMat[numSupp][k2] = 0;
  }
  printf("\nCost By Factory%d and Exporter%d = %d * %d = %d\n", k1, k2, b, a, g = b * a);
  return g;

}

int main() {
  long p = 0, q;
  /*
   * Take Inputs of Max Importers and Exporters aka (Factories and Exporters)
   * User must enter all values in postive integer format only.
   */
  printf("Enter number of Factories: ");
  scanf("%d", & numFact);
  printf("Enter number of Exporters: ");
  scanf("%d", & numSupp);
  
    for (i = 0; i < numSupp; i++) {
      for (j = 0; j < numFact; j++) {
      printf("Cost for Factory%d and Exporter%d = ", j,i);
      scanf("%d", & storMat[i][j]);
    }
    printf("Stock with Exporter%d = ", i);
    scanf("%d", & storMat[i][j]);
  }
  for (j = 0; j < numFact; j++) {
	printf("Needed by Factory%d =", j);
    scanf("%d", & storMat[numSupp][j]);
  }
  printf("\n");
  printr();

  
/*
 * At this step all values are taken as input and output is displayed
 * This while loop performers Iteration on min Function and also checker function as it's condition.
 */
 while (checker() > 1) {
    q = checkMin();
    printf("p= %ld + %ld = ", p, q);
    printf("%ld\n", p += q);
    printr();
  }

  printf("\nFinal Estimated Cost: %ld", p);

  getch();
  return 0;
}