Perl, 103 + 2 = 105 octets
s/$/$"x y===c/gem;$a=$_;$_.=$"while$a=~s/^./!($_.=$&)/gem;s/$1/-/g,$b="$&$b"while/\s(\w)(\1|-)+ /;say$b
Exécuter avec -n0(pénalité de 2 octets).
Explication:
# -n0: read entire input into `$_` at start of program
# (technically speaking it reads to the first NUL byte, but there aren't any)
# We want to be able to extract columns from the input, so we need to add spaces
# to the ends of each line such that each column is complete. Adding too many
# space is OK, so to ensure we have enough, we add a number of spaces equal to the
# length of the input.
s/$/ # At the end of {something},
$" x # append a number of spaces ($" is a space by default)
y===c # obtained by counting the characters in $_
/gem; # where {something} is each (g) line (m)
$a = $_; # store a copy of the transformed input in $a
# The next step is to create a transposition of the input. To do that, we
# repeatedly extract the first column of $a and append it to $_. This will lead to
# a bunch of junk whitespace at the end of $_ (of varying lengths, because once a
# line is empty it's omitted from the extracted column), but we're OK with that.
# To transpose properly, we'd want to place newlines between the extracted
# columns; however, it happens that the rest of the program treats space the same
# way it would newline, and separating via spaces is shorter, so we do that.
while ( # keep looping as long as there are matches
$a =~ s/^./ # replace the first character of {something related to $a}
!( # with the null string (NOT of something truthy)
$_.=$&) # but append that character ($&) to $_
/gem) { # {something} is each (g) line (m) of $a
$_.=$" # append a space ($", equivalent to newline here) to $_
}
# Finally, we repeatedly replace every character in the topmost line with the -
# character (treating a line as continuous through the - character but not through
# other characters), thus finding the lines from top to bottom. Because we
# appended the transpose of $_ to $_ above, each line appears twice: once
# horizontally, once vertically. We find only the horizontal copy, but replace
# both with hyphens.
# (Note: I rewrote the regex into a bit more readable of a form in this ungolfed
# version, because the original version wouldn't allow me room to write comments
# inside it. The two should be equivalent; I tested the golfed version.)
while ( # keep looping as long as there are matches
/\s(\w) # match a space or newline, $1 (a letter/digit/underscore),
(\1|-)+ # any positive number of $1s and hyphens,
\ /x) { # and a space
s/$1/-/g, # changes all $1s to spaces; set $& to $1, $1 becomes invalid
$b = "$&$b" # prepend $& to $b
}
# We need to output the lines from first (i.e. bottom) to last (i.e. top).
# We found them in the opposite order, but reversed them via prepending
# (not appending) the partial results to $b.
say $b # output $b
Une légère subtilité vient ici avec une entrée comme celle-ci:
abc
DDDDDDDDD
abc
abc
abc
Regardez la quatrième ligne ici. Si l'ordre d'écriture était BACBD, il pourrait y avoir véritablement une ligne horizontale de Bs sans violer aucune des hypothèses du problème (à part cela, il n'y a qu'une seule ligne de chaque couleur, ce que nous ne vérifions pas). Afin de contourner cela, nous nous assurons dans la dernière expression régulière que chaque ligne commence par une lettre (ou un chiffre ou un trait de soulignement, mais ceux-ci sont impossibles), et nous comptons sur le fait que les lignes parallèles se trouveront de gauche à droite et en haut -en bas (parce que l'expression régulière trouvera la première correspondance dans la chaîne). En tant que tel, le premier caractère de chaque ligne ambiguë est écrasé avant que la ligne elle-même ne soit considérée comme une correspondance, ce qui empêche la correspondance d'expression régulière.