#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <math.h>
#include <limits.h>

#include "nmglobal.h"
#include "nbench1.h"


/*************************
** ASSIGNMENT ALGORITHM **
*************************/

/*
** DEFINES
*/

#define ASSIGNROWS 101L
#define ASSIGNCOLS 101L

/*
** TYPEDEFS
*/
typedef struct {
	union {
		long *p;
		long (*ap)[ASSIGNROWS][ASSIGNCOLS];
	} ptrs;
} longptr;

/*
** PROTOTYPES
*/
static unsigned long DoAssignIteration(long *arraybase,
		unsigned long numarrays);
static void LoadAssignArrayWithRand(long *arraybase,
		unsigned long numarrays);
static void LoadAssign(long arraybase[][ASSIGNCOLS]);
static void CopyToAssign(long arrayfrom[][ASSIGNCOLS],
		long arrayto[][ASSIGNCOLS]);
static void Assignment(long arraybase[][ASSIGNCOLS]);
static void calc_minimum_costs(long tableau[][ASSIGNCOLS]);
static int first_assignments(long tableau[][ASSIGNCOLS],
		short assignedtableau[][ASSIGNCOLS]);
static void second_assignments(long tableau[][ASSIGNCOLS],
		short assignedtableau[][ASSIGNCOLS]);

/*************
** DoAssign **
**************
** Perform an assignment algorithm.
** The algorithm was adapted from the step by step guide found
** in "Quantitative Decision Making for Business" (Gordon,
**  Pressman, and Cohn; Prentice-Hall)
**
**
** NOTES:
** 1. Even though the algorithm distinguishes between
**    ASSIGNROWS and ASSIGNCOLS, as though the two might
**    be different, it does presume a square matrix.
**    I.E., ASSIGNROWS and ASSIGNCOLS must be the same.
**    This makes for some algorithmically-correct but
**    probably non-optimal constructs.
**
*/
void DoAssign(void)
{
AssignStruct *locassignstruct = &global_assignstruct;  /* Local structure ptr */
long *arraybase = NULL;
char *context = "CPU:Assignment";
unsigned long accumtime;
double iterations;

/*
** See if we need to do self adjustment code.
*/
if(locassignstruct->adjust==0)
{
	/*
	** Self-adjustment code.  The system begins by working on 1
	** array.  If it does that in no time, then two arrays
	** are built.  This process continues until
	** enough arrays are built to handle the tolerance.
	*/
	locassignstruct->numarrays=1;
	while(1)
	{
                arraybase = realloc(arraybase, sizeof(long) * ASSIGNROWS * ASSIGNCOLS * locassignstruct->numarrays);
		if (!arraybase) {
                        fprintf(stderr, "Error in %s, could not allocate memory. Exitting...\n", context);
                        exit(1);
		}

		/*
		** Do an iteration of the assignment alg.  If the
		** elapsed time is less than or equal to the permitted
		** minimum, then allocate for more arrays and
		** try again.
		*/
		if(DoAssignIteration(arraybase,
			locassignstruct->numarrays)>global_min_ticks)
			break;          /* We're ok...exit */

		locassignstruct->numarrays++;
	}
} else {
        arraybase = malloc(sizeof(long) * ASSIGNROWS * ASSIGNCOLS * locassignstruct->numarrays);
	if (!arraybase) {
                fprintf(stderr, "Error in %s, could not allocate memory. Exitting...\n", context);
                exit(1);
	}
}

/*
** All's well if we get here.  Do the tests.
*/
accumtime=0L;
iterations=(double)0.0;

do {
	accumtime+=DoAssignIteration(arraybase,
		locassignstruct->numarrays);
	iterations+=(double)1.0;
} while(TicksToSecs(accumtime)<locassignstruct->request_secs);

/*
** Clean up, calculate results, and go home.  Be sure to
** show that we don't have to rerun adjustment code.
*/
free(arraybase);

locassignstruct->iterspersec=iterations *
	(double)locassignstruct->numarrays / TicksToFracSecs(accumtime);

if(locassignstruct->adjust==0)
	locassignstruct->adjust=1;

return;

}

/**********************
** DoAssignIteration **
***********************
** This routine executes one iteration of the assignment test.
** It returns the number of ticks elapsed in the iteration.
*/
static unsigned long DoAssignIteration(long *arraybase,
	unsigned long numarrays)
{
longptr abase;                  /* local pointer */
unsigned long elapsed;          /* Elapsed ticks */
unsigned long i;

/*
** Set up local pointer
*/
abase.ptrs.p=arraybase;

/*
** Load up the arrays with a random table.
*/
LoadAssignArrayWithRand(arraybase,numarrays);

/*
** Start the stopwatch
*/
elapsed=StartStopwatch();

/*
** Execute assignment algorithms
*/
for(i=0;i<numarrays;i++)
{       /* abase.ptrs.p+=i*ASSIGNROWS*ASSIGNCOLS; */
        /* Fixed  by Eike Dierks */
	Assignment(*abase.ptrs.ap);
	abase.ptrs.p+=ASSIGNROWS*ASSIGNCOLS;
}

/*
** Get elapsed time
*/
return(StopStopwatch(elapsed));
}

/****************************
** LoadAssignArrayWithRand **
*****************************
** Load the assignment arrays with random numbers.  All positive.
** These numbers represent costs.
*/
static void LoadAssignArrayWithRand(long *arraybase,
	unsigned long numarrays)
{
longptr abase,abase1;   /* Local for array pointer */
unsigned long i;

/*
** Set local array pointer
*/
abase.ptrs.p=arraybase;
abase1.ptrs.p=arraybase;

/*
** Set up the first array.  Then just copy it into the
** others.
*/
LoadAssign(*(abase.ptrs.ap));
if(numarrays>1)
	for(i=1;i<numarrays;i++)
	  {     /* abase1.ptrs.p+=i*ASSIGNROWS*ASSIGNCOLS; */
	        /* Fixed  by Eike Dierks */
	        abase1.ptrs.p+=ASSIGNROWS*ASSIGNCOLS;
		CopyToAssign(*(abase.ptrs.ap),*(abase1.ptrs.ap));
	}

return;
}

/***************
** LoadAssign **
****************
** The array given by arraybase is loaded with positive random
** numbers.  Elements in the array are capped at 5,000,000.
*/
static void LoadAssign(long arraybase[][ASSIGNCOLS])
{
unsigned short i,j;

/*
** Reset random number generator so things repeat.
*/
/* randnum(13L); */
randnum(13);

for(i=0;i<ASSIGNROWS;i++)
  for(j=0;j<ASSIGNROWS;j++){
    /* arraybase[i][j]=abs_randwc(5000000L);*/
    arraybase[i][j]=abs_randwc((int32_t)5000000);
  }

return;
}

/*****************
** CopyToAssign **
******************
** Copy the contents of one array to another.  This is called by
** the routine that builds the initial array, and is used to copy
** the contents of the intial array into all following arrays.
*/
static void CopyToAssign(long arrayfrom[ASSIGNROWS][ASSIGNCOLS],
		long arrayto[ASSIGNROWS][ASSIGNCOLS])
{
unsigned short i,j;

for(i=0;i<ASSIGNROWS;i++)
	for(j=0;j<ASSIGNCOLS;j++)
		arrayto[i][j]=arrayfrom[i][j];

return;
}

/***************
** Assignment **
***************/
static void Assignment(long arraybase[][ASSIGNCOLS])
{
short assignedtableau[ASSIGNROWS][ASSIGNCOLS];

/*
** First, calculate minimum costs
*/
calc_minimum_costs(arraybase);

/*
** Repeat following until the number of rows selected
** equals the number of rows in the tableau.
*/
while(first_assignments(arraybase,assignedtableau)!=ASSIGNROWS)
{         second_assignments(arraybase,assignedtableau);
}

#ifdef DEBUG
{
	int i,j;
	printf("\nColumn choices for each row\n");
	for(i=0;i<ASSIGNROWS;i++)
	{
	        printf("R%03d: ",i);
		for(j=0;j<ASSIGNCOLS;j++)
			if(assignedtableau[i][j]==1)
				printf("%03d ",j);
	}
}
#endif

return;
}

/***********************
** calc_minimum_costs **
************************
** Revise the tableau by calculating the minimum costs on a
** row and column basis.  These minima are subtracted from
** their rows and columns, creating a new tableau.
*/
static void calc_minimum_costs(long tableau[][ASSIGNCOLS])
{
unsigned short i,j;              /* Index variables */
long currentmin;        /* Current minimum */
/*
** Determine minimum costs on row basis.  This is done by
** subtracting -- on a row-per-row basis -- the minum value
** for that row.
*/
for(i=0;i<ASSIGNROWS;i++)
{
	currentmin = LONG_MAX;  /* Initialize minimum */
	for(j=0;j<ASSIGNCOLS;j++)
		if(tableau[i][j]<currentmin)
			currentmin=tableau[i][j];

	for(j=0;j<ASSIGNCOLS;j++)
		tableau[i][j]-=currentmin;
}

/*
** Determine minimum cost on a column basis.  This works
** just as above, only now we step through the array
** column-wise
*/
for(j=0;j<ASSIGNCOLS;j++)
{
	currentmin = LONG_MAX;  /* Initialize minimum */
	for(i=0;i<ASSIGNROWS;i++)
		if(tableau[i][j]<currentmin)
			currentmin=tableau[i][j];

	/*
	** Here, we'll take the trouble to see if the current
	** minimum is zero.  This is likely worth it, since the
	** preceding loop will have created at least one zero in
	** each row.  We can save ourselves a few iterations.
	*/
	if(currentmin!=0)
		for(i=0;i<ASSIGNROWS;i++)
			tableau[i][j]-=currentmin;
}

return;
}

/**********************
** first_assignments **
***********************
** Do first assignments.
** The assignedtableau[] array holds a set of values that
** indicate the assignment of a value, or its elimination.
** The values are:
**      0 = Item is neither assigned nor eliminated.
**      1 = Item is assigned
**      2 = Item is eliminated
** Returns the number of selections made.  If this equals
** the number of rows, then an optimum has been determined.
*/
static int first_assignments(long tableau[][ASSIGNCOLS],
		short assignedtableau[][ASSIGNCOLS])
{
unsigned short i,j,k;                   /* Index variables */
unsigned short numassigns;              /* # of assignments */
unsigned short totnumassigns;           /* Total # of assignments */
unsigned short numzeros;                /* # of zeros in row */
int selected=0;                 /* Flag used to indicate selection */

/*
** Clear the assignedtableau, setting all members to show that
** no one is yet assigned, eliminated, or anything.
*/
for(i=0;i<ASSIGNROWS;i++)
	for(j=0;j<ASSIGNCOLS;j++)
		assignedtableau[i][j]=0;

totnumassigns=0;
do {
	numassigns=0;
	/*
	** Step through rows.  For each one that is not currently
	** assigned, see if the row has only one zero in it.  If so,
	** mark that as an assigned row/col.  Eliminate other zeros
	** in the same column.
	*/
	for(i=0;i<ASSIGNROWS;i++)
	{       numzeros=0;
		for(j=0;j<ASSIGNCOLS;j++)
			if(tableau[i][j]==0L)
				if(assignedtableau[i][j]==0)
				{       numzeros++;
					selected=j;
				}
		if(numzeros==1)
		{       numassigns++;
			totnumassigns++;
			assignedtableau[i][selected]=1;
			for(k=0;k<ASSIGNROWS;k++)
				if((k!=i) &&
				   (tableau[k][selected]==0))
					assignedtableau[k][selected]=2;
		}
	}
	/*
	** Step through columns, doing same as above.  Now, be careful
	** of items in the other rows of a selected column.
	*/
	for(j=0;j<ASSIGNCOLS;j++)
	{       numzeros=0;
		for(i=0;i<ASSIGNROWS;i++)
			if(tableau[i][j]==0L)
				if(assignedtableau[i][j]==0)
				{       numzeros++;
					selected=i;
				}
		if(numzeros==1)
		{       numassigns++;
			totnumassigns++;
			assignedtableau[selected][j]=1;
			for(k=0;k<ASSIGNCOLS;k++)
				if((k!=j) &&
				   (tableau[selected][k]==0))
					assignedtableau[selected][k]=2;
		}
	}
	/*
	** Repeat until no more assignments to be made.
	*/
} while(numassigns!=0);

/*
** See if we can leave at this point.
*/
if(totnumassigns==ASSIGNROWS) return(totnumassigns);

/*
** Now step through the array by row.  If you find any unassigned
** zeros, pick the first in the row.  Eliminate all zeros from
** that same row & column.  This occurs if there are multiple optima...
** possibly.
*/
for(i=0;i<ASSIGNROWS;i++)
{       selected=-1;
	for(j=0;j<ASSIGNCOLS;j++)
		if((tableau[i][j]==0L) &&
		   (assignedtableau[i][j]==0))
		{       selected=j;
			break;
		}
	if(selected!=-1)
	{       assignedtableau[i][selected]=1;
		totnumassigns++;
		for(k=0;k<ASSIGNCOLS;k++)
			if((k!=selected) &&
			   (tableau[i][k]==0L))
				assignedtableau[i][k]=2;
		for(k=0;k<ASSIGNROWS;k++)
			if((k!=i) &&
			   (tableau[k][selected]==0L))
				assignedtableau[k][selected]=2;
	}
}

return(totnumassigns);
}

/***********************
** second_assignments **
************************
** This section of the algorithm creates the revised
** tableau, and is difficult to explain.  I suggest you
** refer to the algorithm's source, mentioned in comments
** toward the beginning of the program.
*/
static void second_assignments(long tableau[][ASSIGNCOLS],
		short assignedtableau[][ASSIGNCOLS])
{
int i,j;                                /* Indexes */
short linesrow[ASSIGNROWS];
short linescol[ASSIGNCOLS];
long smallest;                          /* Holds smallest value */
unsigned short numassigns;                      /* Number of assignments */
unsigned short newrows;                         /* New rows to be considered */
/*
** Clear the linesrow and linescol arrays.
*/
for(i=0;i<ASSIGNROWS;i++)
	linesrow[i]=0;
for(i=0;i<ASSIGNCOLS;i++)
	linescol[i]=0;

/*
** Scan rows, flag each row that has no assignment in it.
*/
for(i=0;i<ASSIGNROWS;i++)
{       numassigns=0;
	for(j=0;j<ASSIGNCOLS;j++)
		if(assignedtableau[i][j]==1)
		{       numassigns++;
			break;
		}
	if(numassigns==0) linesrow[i]=1;
}

do {

	newrows=0;
	/*
	** For each row checked above, scan for any zeros.  If found,
	** check the associated column.
	*/
	for(i=0;i<ASSIGNROWS;i++)
	{       if(linesrow[i]==1)
			for(j=0;j<ASSIGNCOLS;j++)
				if(tableau[i][j]==0)
					linescol[j]=1;
	}

	/*
	** Now scan checked columns.  If any contain assigned zeros, check
	** the associated row.
	*/
	for(j=0;j<ASSIGNCOLS;j++)
		if(linescol[j]==1)
			for(i=0;i<ASSIGNROWS;i++)
				if((assignedtableau[i][j]==1) &&
					(linesrow[i]!=1))
				{
					linesrow[i]=1;
					newrows++;
				}
} while(newrows!=0);

/*
** linesrow[n]==0 indicate rows covered by imaginary line
** linescol[n]==1 indicate cols covered by imaginary line
** For all cells not covered by imaginary lines, determine smallest
** value.
*/
smallest=LONG_MAX;
for(i=0;i<ASSIGNROWS;i++)
	if(linesrow[i]!=0)
		for(j=0;j<ASSIGNCOLS;j++)
			if(linescol[j]!=1)
				if(tableau[i][j]<smallest)
					smallest=tableau[i][j];

/*
** Subtract smallest from all cells in the above set.
*/
for(i=0;i<ASSIGNROWS;i++)
	if(linesrow[i]!=0)
		for(j=0;j<ASSIGNCOLS;j++)
			if(linescol[j]!=1)
				tableau[i][j]-=smallest;

/*
** Add smallest to all cells covered by two lines.
*/
for(i=0;i<ASSIGNROWS;i++)
	if(linesrow[i]==0)
		for(j=0;j<ASSIGNCOLS;j++)
			if(linescol[j]==1)
				tableau[i][j]+=smallest;

return;
}