#!/usr/bin/perl

open IN, "<".$ARGV[0] or die "Cannot open $1";
while (<IN>){
    $linea++;
    chomp;
    if ($linea ==1){
	if (/MAX/ || /max/ || /Max/){
	    $tipo = 1;
	}
    }else{
	@lin = split / / ;
	$num=0;
	for $i (@lin){
	    $num++;
	    $costes[$linea-1][$num] = $i;
	}
	$N = $num;
    }
}
close IN;
if ($N != $linea-1) { die "Not a square matrix"; }
print "$N\n";
for $i (1..$N) {
    for $j (1..$N) {
	print "$costes[$i][$j] ";
    }
    print "\n";
}

for $i (1..$N){
    for $j (1..$N){
	if ($tipo==1){
	    $A[$i][$j]=-$costes[$i][$j];
	}else{
	    $A[$i][$j]=$costes[$i][$j];
	}
    }
}


# SUBROUTINE ASSCT ( N, A, C, T )                                         10
#      INTEGER A(130,131), C(130), CH(130), LC(130), LR(130),
#     *        LZ(130), NZ(130), RH(131), SLC(130), SLR(130),
#     *        U(131)
#      INTEGER H, Q, R, S, T
#      EQUIVALENCE (LZ,RH), (NZ,CH)
# C
# C THIS SUBROUTINE SOLVES THE SQUARE ASSIGNMENT PROBLEM
# C THE MEANING OF THE INPUT PARAMETERS IS
# C N = NUMBER OF ROWS AND COLUMNS OF THE COST MATRIX, WITH
# C     THE CURRENT DIMENSIONS THE MAXIMUM VALUE OF N IS 130
# C A(I,J) = ELEMENT IN ROW I AND COLUMN J OF THE COST MATRIX
# C ( AT THE END OF COMPUTATION THE ELEMENTS OF A ARE CHANGED)
# C THE MEANING OF THE OUTPUT PARAMETERS IS
# C C(J) = ROW ASSIGNED TO COLUMN J (J=1,N)
# C T = COST OF THE OPTIMAL ASSIGNMENT
# C ALL PARAMETERS ARE INTEGER
# C THE MEANING OF THE LOCAL VARIABLES IS
# C A(I,J) = ELEMENT OF THE COST MATRIX IF A(I,J) IS POSITIVE,
# C          COLUMN OF THE UNASSIGNED ZERO FOLLOWING IN ROW I
# C          (I=1,N) THE UNASSIGNED ZERO OF COLUMN J (J=1,N)
# C          IF A(I,J) IS NOT POSITIVE
# C A(I,N+1) = COLUMN OF THE FIRST UNASSIGNED ZERO OF ROW I
# C            (I=1,N)
# C CH(I) = COLUMN OF THE NEXT UNEXPLORED AND UNASSIGNED ZERO
# C         OF ROW I (I=1,N)
# C LC(J) = LABEL OF COLUMN J (J=1,N)
# C LR(I) = LABEL OF ROW I (I=1,N)
# C LZ(I) = COLUMN OF THE LAST UNASSIGNED ZERO OF ROW I(I=1,N)
# C NZ(I) = COLUMN OF THE NEXT UNASSIGNED ZERO OF ROW I(I=1,N)
# C RH(I) = UNEXPLORED ROW FOLLOWING THE UNEXPLORED ROW I
# C         (I=1,N)
# C RH(N+1) = FIRST UNEXPLORED ROW
# C SLC(K) = K-TH ELEMENT CONTAINED IN THE SET OF THE LABELLED
# C          COLUMNS
# C SLR(K) = K-TH ELEMENT CONTAINED IN THE SET OF THE LABELLED
# C          ROWS
# C U(I) = UNASSIGNED ROW FOLLOWING THE UNASSIGNED ROW I
# C        (I=1,N)
# C U(N+1) = FIRST UNASSIGNED ROW
# C
# C THE VECTORS C,CH,LC,LR,LZ,NZ,SLC,SLR MUST BE DIMENSIONED
# C AT LEAST AT (N), THE VECTORS RH,U AT  LEAST AT (N+1),
# C THE MATRIX A AT LEAST AT (N,N+1)
# C
# C INITIALIZATION

$MAXNUM = 10**14;
$NP1 = $N+1;
for $j (1..$N){
    $C[$J] = 0;
    $LZ[$J] = 0;
    $NZ[$J] = 0;
    $U[$J] = 0;
}
$U[$NP1] = 0;
$T = 0;
# C REDUCTION OF THE INITIAL COST MATRIX
for $J (1..$N){
    $S = $A[1][$J];
    for $L (2 .. $N) {
	if ( $A[$L][$J] < $S ){  $S = $A[$L][$J]; }
    }
    $T += $S;
    for $I (1..$N){
	$A[$I][$J] -= $S;
    }
}
for $I (1..$N){
    $Q = $A[$I][1];
    for $L (2..$N){
	if  ( $A[$I][$L] < $Q ) { $Q = $A[$I][$L]; }
    }
    $T += $Q;
    $L = $NP1;
    for $J (1..$N){
	$A[$I][$J] -= $Q;
	next if ( $A[$I][$J] != 0 );
	$A[$I][$L] = -$J;
	$L = $J;
    }
}
# C CHOICE OF THE INITIAL SOLUTION
$K = $NP1;
for  $I (1..$N){
        $LJ = $NP1;
        $J = -$A[$I][$NP1];
a80:
      if ( $C[$J] == 0 ) { goto b130; }
        $LJ = $J;
        $J = -$A[$I][$J];
        if ( $J != 0 ) { goto a80; }
        $LJ = $NP1;
        $J = -$A[$I][$NP1];
c90:   $R = $C[$J];
        $LM = $LZ[$R];
        $M = $NZ[$R];
   while ( $M != 0 ) {
        if ( $C[$M] == 0 ) { goto d120; }
        $LM = $M;
        $M = -$A[$R][$M];
      }
    
e110:   $LJ = $J;
        $J = -$A[$I][$J];
        if ( $J != 0 ) { goto c90; }
        $U[$K] = $I;
        $K = $I;
    next;

d120:   $NZ[$R] = -$A[$R][$M];
        $LZ[$R] = $J;
        $A[$R][$LM] = -$J;
        $A[$R][$J] = $A[$R][$M];
        $A[$R][$M] = 0;
        $C[$M] = $R;
    
b130:   $C[$J] = $I;
        $A[$I][$LJ] = $A[$I][$J];
        $NZ[$I] = -$A[$I][$J];
        $LZ[$I] = $LJ;
    $A[$I][$J] = 0;
}

# C RESEARCH OF A NEW ASSIGNMENT
res: ; 
 if ( $U[$NP1] == 0 ){ &output(); exit; } 
for $I (1..$N){
    $CH[$I] = 0;
    $LC[$I] = 0;
    $LR[$I] = 0;
    $RH[$I] = 0;
}
      $RH[$NP1] = -1;
      $KSLC = 0;
      $KSLR = 1;
      $R = $U[$NP1];
      $LR[$R] = -1;
      $SLR[1] = $R;

if ( $A[$R][$NP1] != 0 ) {
ERR170:  $L = -$A[$R][$NP1];
      if ( $A[$R][$L] == 0 || $RH[$R] != 0 ) { goto f180; }
      $RH[$R] = $RH[$NP1];
      $CH[$R] = -$A[$R][$L];
      $RH[$NP1] = $R;
f180: if ( $LC[$L] == 0 ) { goto g200; }
      if ( $RH[$R] == 0 ) { goto h210; }
i190: $L = $CH[$R];
      $CH[$R] = -$A[$R][$L];
      if ( $A[$R][$L] != 0 ) { goto f180; }
      $RH[$NP1] = $RH[$R];
      $RH[$R] = 0;
      goto f180;
	  
g200: $LC[$L] = $R;
      if ( $C[$L] == 0 ) { goto z360; }
      $KSLC++;
      $SLC[$KSLC] = $L;
      $R = $C[$L];
      $LR[$R] = $L;
      $KSLR++;
      $SLR[$KSLR] = $R;
	  
      if ( $A[$R][$NP1] != 0 ) { goto ERR170; }
h210: ;  # CONTINUE
      if ( $RH[$NP1] > 0 ) { goto z350; }
}

# C REDUCTION OF THE CURRENT COST MATRIX
z220:  $H = $MAXNUM;
     for  $J (1..$N){
	 next if ( $LC[$J] != 0 );
	 for $K (1..$KSLR) {
	     $I = $SLR[$K];
          if ( $A[$I][$J] < $H ) { $H = $A[$I][$J]; }
	 }
     }
$T += $H;
for $J (1..$N) { 
    next if ( $LC[$J] != 0 );
    for $K (1 .. $KSLR) {
	$I = $SLR[$K];
	$A[$I][$J] = $A[$I][$J]-$H;
	next if ( $A[$I][$J] != 0 );
	if ( $RH[$I] == 0 ){
	    $RH[$I] = $RH[$NP1];
	    $CH[$I] = $J;
	    $RH[$NP1] = $I;
	}
	$L = $NP1;
z260:     $NL = -$A[$I][$L];
          if ( $NL == 0 ) { goto z270; }
          $L = $NL;
          goto z260;
z270:     $A[$I][$L] = -$J;
    }
}

if ( $KSLC != 0 ) {
    for $I (1..$N) { 
	next if ( $LR[$I] != 0 );
	for $K (1..$KSLC) {
	    $J = $SLC[$K];
	    if ( $A[$I][$J] <= 0 ) {
		$L = $NP1;
z300:     $NL = - $A[$I][$L];
          if ( $NL == $J ) { goto z310; }
          $L = $NL;
          goto z300;
z310:     $A[$I][$L] = $A[$I][$J];
		$A[$I][$J] = $H;
		next;
	    }
	    $A[$I][$J] = $A[$I][$J]+$H;
	}
    }
}
z350:  $R = $RH[$NP1];
      goto i190;

# C ASSIGNMENT OF A NEW ROW
z360: $C[$L] = $R;
      $M = $NP1;
z370:  $NM = -$A[$R][$M];
      if ( $NM == $L ) { goto z380; }
      $M = $NM;
      goto z370;
z380:  $A[$R][$M] = $A[$R][$L];
      $A[$R][$L] = 0;
      if ( $LR[$R] >= 0 ) {
	  $L = $LR[$R];
	  $A[$R][$L] = $A[$R][$NP1];
	  $A[$R][$NP1] = -$L;
	  $R = $LC[$L];
	  goto z360;
      }
z390: $U[$NP1] = $U[$R];
    $U[$R] = 0;
  &output;
  goto res;
exit ;


sub output {
#    for $i (1..$N){
#	for $j (1..$N){
#	    print "$A[$i][$j] ";
#	}
#	print "\n";
#   }
    print "\nRows assigned to columns:\n@C\n";
    $Z=0;
    for $i (1..$N){
	$Z += $costes[$i][$C[$i]];
    }
    print "Value of the objective function: $Z\n";
}



    
__END__