Code Golf: Decision Tree

问题

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.

In Google Code Jam 2009, Round 1B, there is a problem called Decision Tree that lent itself to rather creative solutions.

Post your shortest solution; I'll update the Accepted Answer to the current shortest entry on a semi-frequent basis, assuming you didn't just create a new language just to solve this problem. :-P

Current rankings:

• 107 Perl
• 121 PostScript (binary)
• 132 Ruby
• 154 Arc
• 160 PostScript (ASCII85)
• 170 PostScript
• 192 Python
• 196 JavaScript
• 199 Common Lisp
• 212 LilyPond
• 273 Scheme
• 280 R
• 281 sed w/ bc
• 312 Haskell
• 314 PHP
• 339 m4 w/ bc
• 346 C
• 381 Fortran
• 462 Java
• 718 OCaml
• 759 F#
• 1554 sed
• C++ not qualified for now

回答1:

Perl in 107 characters

say("Case #$_:"),
$_=eval"''".'.<>'x<>,
s:[a-z]+:*(/ $&\\s/?:g,s/\)\s*\(/):/g,
eval"\$_=<>;say$_;"x<>for 1..<>

Newlines for legibility; none of them is necessary or counted in.

It uses features found only in the latest versions of Perl, so run with perl -M5.010 or later.

I used to be a Perl noob too, so this works almost the same as the ruby one. Original version 126 chars, optimizations by peutri.

Backlinks:

Word Aligned - Power Programming

回答2:

sed in 1554 chars (pure) / 281 (with bc)

Yes, seriously.

Usage: sed -r -f thisfile.sed < input.in > output.out

(works on GNU sed)

1d
/ /!{x
s/^$/Case #Y:/
:i
s/9Y/Y0/
ti
s/#Y/#0Y/
s/:/:0123456789/
s/(.)Y(.*):[0-9]*\1(.).*/\3\2Y:/
x
G
s/.*\n|Y//gp
z
:p
N
/[()]/s/ |\n//g
y/()/JK/
tp
H
d}
G
s/\n[^J]*/ %/
s/[^JK]*$//
:c
s/J1?([.-9]+)(.*)K/\2@\1/
/%@/by
:b
/J/s/T//
s/J([^JK]*)K/TC\1B/
tb
/ (.+) .*%\1C/{s/%[^C]*/%/
s/T.*B//
by}
s/%.*T/%/
:y
y/CB/JK/
tc
s/.\.0*\b//g
:r
/@.*@/{s/\w*@\w*$/C&B/
s/C(\w)(.*B)/\1C\2~/
s/"[^"]*/&0/g
:t
s/(\w)(C.*)(\w)B(.*~)/\1\2B\3\4\1\3/
T
s/~(10|2[01]|3[0-2]|4[0-3]|5[0-4]|6[0-5]|7[0-6]|8[0-7]|9.)/&Q/
s/(.)(.)Q/\2\1/
s/~0\w/`00/
s/~1\B/`0/
s/~22/`04/
s/~23/`06/
s/~24/`08/
s/~33/`09/
s/~25/`10/
s/~26|~34/`12/
s/~27/`14/
s/~28|~44/`16/
s/~29|~36/`18/
s/~35/`15/
s/~45/`20/
s/~37/`21/
s/~38|~46/`24/
s/~55/`25/
s/~39/`27/
s/~47/`28/
s/~56/`30/
s/~48/`32/
s/~57/`35/
s/~49|~66/`36/
s/~58/`40/
s/~67/`42/
s/~59/`45/
s/~68/`48/
s/~77/`49/
s/~69/`54/
s/~78/`56/
s/~79/`63/
s/~88/`64/
s/~89/`72/
s/~99/`81/
s/`(.)(.)/~\1'\2/
bt
:
s/(~.)'/\1/
s/..'/K&/
/K/bk
:v
s/=(,?.)'/\1/
s/,/1'/
t
s/B(.*)~/\1B"/
tr
s/"(\w*)0/A\1/g
/A.*A/{s/A[^A]*$/J&K/
:k
s/([^A])(J.*)([^A])K/\2K\1\3/
s/K(10|2[01]|3[0-2]|4[0-3]|5[0-4]|6[0-5]|7[0-6]|8[^9]|9.)/&Q/
s/(.)(.)Q/\2\1/
s/K0/=/
s/K11/=2/
s/K12/=3/
s/K13|K22/=4/
s/K14|K23/=5/
s/K15|K24|K33/=6/
s/K16|K25|K34/=7/
s/K(17|26|35|44)/=8/
s/K(18|27|36|45)/=9/
s/K(19|28|37|46|55)/W0/
s/K(29|38|47|56)/W1/
s/K(39|48|57|66)/W2/
s/K49|K58|K67/W3/
s/K59|K68|K77/W4/
s/K69|K78/W5/
s/K79|K88/W6/
s/K89/W7/
s/K99/W8/
s/W/=,/
/'/bv
s/\b=/K:/
tk
s/[:JK]A?//g
s/,/,0123456789GF/
s/(.),.*\1(.).*F/\2/
s/G/,0/
tk}
/A.*A/bv}
s/\w*C.*A//
tr
s/.*@/./

This solution omits the leading zero in front of the decimal point, and does not handle cases where the answer is 1.00. Luckily, the GCJ judge accepts the lack of a zero, and does not have any cases where the answer is 1.00.

To include the leading zero, change the last line to s/.*@/0./; and to handle a 1.00 case, append the line s/^$/1/.

Here's a solution that outsources the multiplication to bc:

1d
/ /!{x
s/\n.*//
s/.*/echo 0&+1|bc/e
x
g
s/.*/Case #&:/p
:p
N
/[()]/s/ |\n//g
y/()/JK/
tp
H
d}
G
s/\n[^J]*/ %/
s/[^JK]*$//
:c
s/J([.-9]+)(.*)K/\2*\1/
/%\*/s/.*%.(.*)/echo \1|bc -l/e
:b
/J/s/T//
s/J([^JK]*)K/TC\1B/
tb
/ (.+) .*%\1C/{s/%[^C]*/%/
s/T.*B//
b}
s/%.*T/%/
:
y/CB/JK/
tc

回答3:

LilyPond: 212 characters

Craziness! Utter ridiculousness!! LilyPond, with its built-in Scheme interpreter, manages to outdo Scheme by more than FIFTY BYTES! Holy acrobatic flying mooses in tights!!

x=#lambda
w=#read
#(letrec((v(x(a)(map a(iota(w)1))))(c(x(f q)(*(car q)(if(any list? q)(c
f((if(memq(cadr q)f)caddr cadddr)q))1)))))(v(x(i)(w)(set! @(w))(format
#t"Case #~a:
~{~y~}"i(v(x i(w)(c(v(x i(w)))@)))))))

Usage: lilypond thisfile.ly <input.in >output.out 2>/dev/null

Credit goes to cky for writing the Scheme solution this was based on, though this version is now substantially different. Seriously, though, the Scheme could be golfed a bit further...

回答4:

PostScript: 170 (regular) / 160 (ASCII85) / 121 (binary)

My shortest (regular) PostScript solution so far, provided that you rename the input file to "r" (170 characters, including newlines); uses a GhostScript-specific procedure (=only):

1[/:{repeat}/!{exch token{\ exch known{/<>}if]pop]]3 index mul
!}if}(]){token pop}/?(r)(r)file([){?]}>>begin
1[{(Case #)2{=only}:(:)=[/|[def[{[/\<<[{[/}:>>def |]! =}:}for

Usage: cp input.in r; gs -q -dNOPROMPT -dNODISPLAY -dBATCH thisfile.ps > output.out

Here's a binary version of this in 121 bytes (backslashes and unprintable characters escaped):

1[/!{\x92>\x92\xab{\\\x92>\x92`\x92p{]\x92u}if]]3\x92X\x92l!}if}(]){\x92\xab\x92u}/r(r)\x928\x92A([){r]}>>\x92\r1[{(Case #)\x92v=only[/:\x928[\x923=[{[/\\<<[{[/}\x92\x83>>\x923:]! =}\x92\x83}\x92H

If characters outside the ASCII printable range are disallowed, PS has built-in ASCII85 encoding of binary sources. We therefore have the following 160-byte solution in all ASCII printable characters:

1[([){r]}/r(r)<~OuSUj0-P\*5*Dsn>`q:6@$5JU?'9>YBkCXV1Qkk'Ca"4@Apl(5.=75YP')1:5*?@0>C.bc@<6!&,:Se!4`>4SH!;p_OuQ[/1Herh>;'5D4Bm/:07B"95!G,c3aEmO4aiKGI?I,~>cvx exec

回答5:

Ruby in 132

Improved by leonid. Newlines are essential.

def j
'1
'..gets
end
j.map{|c|s=j.map{gets}*''
puts"Case #%d:"%c,j.map{gets;eval s.gsub(/[a-z]+/,'*(/ \&\b/?').gsub /\)\s*\(/,'):'}}

Ruby in 136

def j;1..gets.to_i;end;j.map{|c|m=j.map{gets}*"";puts"Case ##{c}:";j.map{gets;p eval m.gsub(/[a-z]+/,'*(/ \0\s/?').gsub /\)\s*\(/,'):'}}

I just learned about *"" being equivalent to .join"". Also realised that map could be used in a few places

Ruby in 150

1.upto(gets.to_i){|c|m=eval("gets+"*gets.to_i+"''");puts"Case ##{c}:";1.upto(gets.to_i){gets;p eval m.gsub(/[a-z]+/,'*(/ \0\s/?').gsub /\)\s*\(/,'):'}}

I am just a noob to ruby, so there is probably still a lot of room for improvement

回答6:

Python in 192

import re;S=re.sub;R=raw_input;I=input;c=0;exec r"c+=1;L=S('\) *\(',')or ',S('([a-z]+)','*(\' \\1 \'in a and',eval(('+R()'*I('Case #%s:\n'%c))[1:])));exec'a=R()+\' \';print eval(L);'*I();"*I()

回答7:

Common Lisp, 199 bytes

Wrapped every 80 characters:

(defun r()(read))(dotimes(i(r))(format t"~&Case #~D:"(1+ i))(r)(set'z(r))(dotime
s(a(r))(r)(print(do((g(mapcar'read(make-list(r))))(p 1(*(pop c)p))(c z(if(find(p
op c)g)(car c)(cadr c))))((not c)p)))))

Spaced and indented:

(defun r () (read))
(dotimes (i (r))
  (format t "~&Case #~D:" (1+ i))
  (r)
  (set 'z (r))
  (dotimes (a (r))
    (r)
    (print
      (do ((g (mapcar 'read (make-list (r))))
           (p 1 (* (pop c) p))
           (c z (if (find (pop c) g)
                    (car c)
                    (cadr c))))
          ((not c) p)))))

回答8:

C - 346 bytes

Compile with gcc -w

#define N{int n=atoi(gets(A));for(;n--;)
T[999];F[99];char*t,*f,*a,A[99];float p(){float
d,m=1;for(;*t++^40;);sscanf(t,"%f %[^ (]",&d,A);if(*A^41){for(f=F;m**f;){for(;*f&&*f++^32;);for(a=A;*a&&*f==*a;f++,a++);m=*a||*f&64;}d*=!m*p()+m*p();}return
d;}main(I)N{printf("Case #%d:\n",I++);t=T;N
for(gets(t);*++t;);}N gets(F),t=T,printf("%f\n",p());}}}

回答9:

Arc, 143 154 characters

Very similar to the CL one, but Arc sure has terse identifiers. Wrapped every 40 chars:

(for i 1((= r read))(prn"Case #"i":")(r)
(= z(r))(repeat(r)(r)(loop(= g(n-of(r)(r
))c z p 1)c(= p(*(pop c)p)c(if(pos(pop c
)g)c.0 cadr.c)))prn.p))

Indented:

(for i 1 ((= r read))
  (prn "Case #" i ":")
  (r)
  (= z (r))
  (repeat (r)
    (r)
    (loop (= g (n-of (r) (r))
             c z
             p 1)
       c
       (= p (* (pop c) p)
          c (if (pos (pop c) g)
                (c 0)
                (cadr c))))
    (prn p)))

Backlink: Word Aligned - Power Programming

回答10:

JavaScript in 196 bytes

r='replace'
q=readline
for(n=0,t=q();t-n++;){for(print('Case #'+n+':'),d='',x=q();x--;d+=q());for(x=q();x--;)print(eval(d[r](/([a-z]+)/g,'*({'+q()[r](/ /g,':1,z')+':1}.z$1?')[r](/\) *\(/g,'):')))}

Usage: $ smjs thisfile.js <input.in

With contributions by Hyperlisk.

回答11:

PHP in 314

<?php function q(){return trim(fgets(STDIN));}for($n=q($x=0);$x++<$n;){for($s=q($t='');$s--;$t.=q());echo"Case #$x:\n";for($z=q();$z--;){$l=explode(' ',q());$l[0]=0;printf("%f\n",eval('return'.preg_replace(array('/\(/','/(\w+),/','/(\d\)*),\((\d)/','/^./'),array(',(','*(in_array("$1",$l,1)?','$1:$2'),$t).';'));}}

回答12:

FORTRAN - 381

Save as a.F95
Compile with f95 a.F95

#define _ ENDDO
#define A READ(t(k:l-1),*),a
#define Q j=1,n;READ"(A)",s
#define R READ*,n;DO
#define S k+SCAN(t(k:),'()')
CHARACTER(999)s,t,u;R i=1,n;t="";PRINT"('Case #'i0':')",i
R Q;t=TRIM(t)//s;_;R Q;d=1;k=1;DO;k=S;l=S-1
IF(t(l:l)>"(")EXIT;A,u;d=d*a;k=l;m=0
IF(INDEX(s," "//TRIM(u)//" ")>0)CYCLE;DO;IF(')'>t(k:k))m=m+2;m=m-1;k=k+1
IF(1>m)EXIT;k=S-1;_;_;A;d=d*a;PRINT*,d;_;_;END

By using the default format, each of the results starts with 2 spaces, but the google judge permits it. Thanks google judge!

EXPANDED VERSION

CHARACTER(999)s,t,u
READ*,n
DO i=1,n
    t=""
    PRINT"('Case #'I0':')",i
    READ*,n
    DO j=1,n
        READ"(A)",s
        t=TRIM(t)//s
    ENDDO
    READ*,n
    DO j=1,n
        READ"(A)",s
        d=1
        k=1
        DO
            k=k+SCAN(t(k:),'()')
            l=k+SCAN(t(k:),'()')-1
            IF(t(l:l)>"(")THEN
                READ(t(k:l-1),*),a
                d=d*a
                PRINT*,d
                EXIT
            ELSE
                READ(t(k:l-1),*),a,u
                d=d*a
                k=l
                m=0
                IF(INDEX(s," "//TRIM(u)//" ")>0)CYCLE
                DO
                    IF(')'>t(k:k))m=m+2
                    m=m-1
                    k=k+1
                    IF(1>m)EXIT
                    k=k+SCAN(t(k:),'()')-1
                ENDDO
            ENDIF
        ENDDO
    ENDDO
ENDDO
END

回答13:

Haskell, 312 characters

Here's another aproach to Haskell. I left the dirty work to the Prelude's lex. The wrapping around it is Text.ParserCombinators.ReadP. Importing it cost 36 characters on its own—ugh!

The parser is a Features -> SExp -> Cuteness function, which spares me most of the type declarations in quibble's/yairchu's solution.

import Text.ParserCombinators.ReadP
main=f(\t->do putStrLn$"Case #"++show t++":";s<-r g;r$print.fst.head.($id=<<s).readP_to_S.d.tail.words=<<g)
d x=do"("<-e;w<-e;c<-do{f<-e;y<-d x;n<-d x;u$if elem f x then y else n}<++u 1.0;e;u$c*read w
f x=do n<-g;mapM x[1..read n]
e=readS_to_P lex
r=f.const
g=getLine
u=return

It used to use Control.Monad's join, forM_ and replicateM, but it turns out it takes less space to redefine them approximately than to import.

I also abandoned the Prelude's readParen in favor of just calling lex before and after. In the current version, there is no need to verify the closing parenthesis: on a valid input it will always be there. On the other hand, it is vital to check the opening one: since the number is only converted after the whole subexpression has been read, a lot of backtracking would be needed to align to the correct parse.

On a theoretical machine with infinite memory and time to spare, the "("<- part might be dropped (4 characters' gain, 308 in total). Unless the call to read just aborts. On mine, the stack just overflows pretty fast.

回答14:

Java in 467 bytes

This uses the javascript interpreter contained in java 6.

import java.util.*;class D{static{Scanner c=new
Scanner(System.in);int n=c.nextInt(),i=0,l;while(i++<n){l=c.nextInt();String
s="(";while(l-->=0)s+=c.nextLine();System.out.println("Case #"+i+":");l=c.nextInt();while(l-->0)try{c.next();System.out.println(new
javax.script.ScriptEngineManager().getEngineByName("js").eval(s.replace(")","))").replaceAll("\\) *\\(",":(").replaceAll("[a-z]+","*(/ $0 /.test('"+c.nextLine()+" ')?")));}catch(Exception
x){}}System.exit(0);}}

Thanks Varan, Chris and pfn (indirectly) for helping me shorten it.
Please see my other (even shorter!) java answer.

回答15:

m4 with echo and bc, 339 bytes

This solution is a complete and utter hack, and it gives me a headache. It contains, among other things, escaped double quotes, unescaped double quotes, unescapable backquote and single quote pairs (including a nested pair seven quotes deep), unquoted regular expressions, outsourcing decimal multiplication to bc, and the use of craZy caSE to circumvent macro expansion. But it had to be done, I guess. :p

This adds an "ultimate macroizing" solution to the previous kinds of solutions (iterated loops, recursion w/ lambda mapping, labels and branches, regexp and eval, etc.)

I think a good term for this is "macroni code" :D

(wrapped every 60 characters, for clarity)

define(T,`translit($@)')define(Q,`patsubst($@)')define(I,0)Q
(T(T(T(Q(Q(Q(Q(Q(Q(T(include(A),(),<>),>\s*>,>>),>\s*<,>;),\
([a-z]+\)\s*<,`*ifElsE<rEgExp<P;``````` \1 ''''''';0>;0;<'),
^<,`defiNe<````I';iNcr<I>>\\"Case `#'I:\\"defiNe<`A'''';'),^
[0-9]*),.+ [0-9]+.*,`dEfiNE<```P';`\& '''>A'),<>;N,`(),n'),E
,e),()),.*,`syscmd(`echo "\&"|bc -l')')

Usage: $ cp input.in A; m4 thisfile.m4 > output.out

I'm an m4 n00b, though, having learned it only an hour before writing this. So there's probably room for improvement.

回答16:

C++ in 698 bytes
Compile with 'g++ -o test source.cpp -include iostream -include vector -include sstream'

#define R(x,f,t) for(int x=f;x<t;x++){
#define S(x) x.size()
#define H string
#define U while
#define I if
#define D cin>>
#define X t.substr(p,S(t))
using namespace std;
int main(){int h,l,n,a,p,Y,W;D h;for(int q=1;q<=h;q++){D l;H s;char c;D c;R(i,0,l)H L;getline(cin,L);R(j,0,S(L))I (L[j]==41||L[j]==40)s+=32;s+=L[j];I(L[j]==40)s+=32;}}D a;printf("Case #%d:\n",q);R(i,0,a)H N;D N;D n;vector<H>f;R(j,0,n)D N;f.push_back(N);}H t=s;float P=1;p=0;U(p<S(t)-1){p=0;U(t[p]!=48&&t[p]!=49)p++;t=X;stringstream T(t);float V;T>>V;H F;T>>F;P*=V;I(F[0]==41)break;Y=0;R(j,0,S(f))if(F==f[j])Y=1;}p=t.find(40)+1;t=X;p=0;I(Y==0){W=1;U (W>0){I(t[p]==40)W++;I(t[p]==41)W--;p++;}t=X;p=0;}}cout<<P<<endl;}}return 0;}

EDIT: I'm sorry; I thought it was ok for the includes (eg, C works even w/o including basic libraries), while I'm sure it would be if I decleared the defines this way. I'm not home now, and I won't be for some time: I won't be able to modify it. Just ignore my submission.

回答17:

OCaml in 718 bytes

I'm an OCaml n00b, so this is probably much longer than it needs to be.

Usage: ocaml thisfile.ml <input.in >output.out

#load"str.cma";;open List;;open String;;open Str;;let x=length and
y=Printf.printf and e=global_replace and h=float_of_string and b=regexp and
k=index and r=read_line and a=read_int and w s m c=sub s(c+1)(m-c-1);;for i=1to
a()do y"Case #%d:\n"i;let t=let n=a()in let rec g d j=if j>n then d else
g(d^(r()))(j+1)in e(b" ")""(e(b"\\b")"^"(g""1))and n=a()in let rec z j=if j>n
then()else let q=tl(split(b" ")(r()))in let rec g l j s p=let o=k s '('and c=k
s ')'in if j then let f=w s c o in if contains f '('then let m=k s '^'in let
c=index_from s(m+1)'^'in g 0(mem(w s c m)q)(w s(x s)c)(h(w s m o)*.p)else h f*.p
else if o<c then g(l+1)j(w s(x s)o)p else g(l-1)(l=1)(w s(x s)c)p in y"%f\n"(g
0(0=0)t 1.);z(j+1)in z 1done

回答18:

Scheme (Guile 1.8)

Here's my version at 278 bytes (with improvements from KirarinSnow to bring it down to 273), after stripping off all the newlines (except ones in string literals, of course). It only works on Guile 1.8 (since in standard Scheme, define is a syntax, not an object, but Guile represents it as an object anyway).

(define ! define)
(!(c f p w . r)(if(null? r)(* p w)(apply c f(* p w)((if(memq(car r)f)cadr caddr)r))))
(!(d . l)(map display l))
(!(r . x)(read))
(! n(r))
(do((i 1(1+ i)))((> i n))(r)(let((t(r)))(d"Case #"i":
")(do((a(r)(1- a)))((= a 0))(r)(d(apply c(map r(iota(r)))1 t)"
"))))

回答19:

Pure java in 440 bytes

A shorter java solution that doesn't use any eval trick. Can be reduced to 425 by removing System.exit(0) if stderr output is ignored.

import java.util.*;enum A{_;Scanner c,d;float p(String a){return
d.nextFloat()*(d.hasNext("\\D+")?a.contains(' '+d.next()+' ')?p(a)+0*p(a):0*p(a)+p(a):1);}{c=new
Scanner(System.in);for(int n=c.nextInt(),i=0,l;i++<n;){String
s="";for(l=c.nextInt();l-->=0;)s+=c.nextLine();System.out.println("Case #"+i+":");for(l=c.nextInt();l-->0;){c.next();d=new
Scanner(s.replaceAll("[()]"," "));System.out.println(p(c.nextLine()+' '));}}System.exit(0);}}

回答20:

Haskell, 514 bytes (I suck?).

Based on quibble's solution:

import Control.Monad
import Text.ParserCombinators.Parsec
data F=N|F String(Float,F)(Float,F)
r=return
f=many1 letter>>= \i->w>>d>>= \t->d>>=r.F i t
d=char '('>>w>>many1(oneOf".0123456789")>>= \g->w>>(f<|>r N)>>= \p->char ')'>>w>>r(read g,p)
w=many$oneOf" \n"
g=getLine
l=readLn
m=replicateM
main=l>>= \n->forM_[1..n]$ \t->putStrLn("Case #"++show t++":")>>l>>=(`m`g)>>=(\(Right q)->l>>=(`m`p q)).parse d"".join
z(p,f)=(p*).y f
y N _=1
y(F n t f)x=z(if n`elem`x then t else f)x
p q=fmap(drop 2.words)g>>=print.z q

回答21:

C in 489 bytes

Code wrapped at 80 chars, there are actually just 3 lines.

Save in a.c and compile with: gcc -w a.c -o a

#define S int I,N;scanf("%d\n",&N);for(I=-1;++I<N;)
#define M 1000
char B[M],Z[M],Q[M]={' '},*F[M],*V;float W[M],H;int J,C,L[M],R[M];t(){V=strtok(0
," \n()");}p(){int U=C++;F[U]=0;if(!V)t();sscanf(V,"%f",W+U);t();if(V&&*V>='a')s
trcpy(Q+1,V),V=0,F[U]=strdup(strcat(Q," ")),L[U]=p(),R[U]=p();return U;}main(){S
{printf("Case #%d:\n",I+1);*B=0;{S strcat(B,gets(Z));}V=strtok(B," \n(");C=0,p()
;{S{strcat(gets(B)," ");for(J=0,H=W[0];F[J];J=strstr(B,F[J])?L[J]:R[J],H*=W[J]);
printf("%f\n",H);};}}}

回答22:

F#: 759 significant chars (Wow, I'm bad at this ;) )

Minimized version

open System.Text.RegularExpressions
type t=T of float*(string*t*t)option
let rec e=function x,T(w,Some(s,a,b))->e(x,if Set.contains s x then a else b)*w|x,T(w,_)->w
let rec h x=Regex.Matches(x, @"\(|\)|\d\.\d+|\S+")|>Seq.cast<Match>|>Seq.map (fun x -> x.Value)|> Seq.toList
let rec p=function ")"::y->p y|"("::w::x::y->match x with ")"->T(float w,None),y|n->let a,f=p y in let b,g=p f in T(float w,Some(n,a,b)),g
let solve input =
    Regex.Matches(input,@"(\(((?<s>\()|[^()]|(?<-s>\)))*\)(?(s)(?!)))\s+\d+\s+((\S+\s\d(.+)?\s*)+)")
    |>Seq.cast<Match>
    |>Seq.map(fun m->fst(p(h(m.Groups.[1].Value))), [for a in m.Groups.[3].Value.Trim().Split([|'\n'|])->set(a.Split([|' '|]))])
    |>Seq.iteri(fun i (r,c)->printfn"Case #%i"(i+1);c|>Seq.iter(fun x->printfn"%.7F"(e(x, r))))

Readable version

open System.Text.RegularExpressions

type decisionTree = T of float * (string * decisionTree * decisionTree) option
let rec eval = function 
    | x, T(w, Some(s, a, b)) -> eval(x, if Set.contains s x then a else b) * w
    | x, T(w, _) -> w

// creates a token stream
let rec tokenize tree =
    Regex.Matches(tree, @"\(|\)|\d\.\d+|\S+")
    |> Seq.cast<Match>
    |> Seq.map (fun x -> x.Value)
    |> Seq.toList

// converts token stream into a decisionTree
let rec parse = function
    | ")"::xs -> parse xs
    | "("::weight::x::xs ->
        match x with
        | ")" -> T(float weight, None), xs
        | name ->
            let t1, xs' = parse xs
            let t2, xs'' = parse xs'
            T(float weight, Some(name, t1, t2)), xs''

// uses regex to transform input file into a Seq<decisionTree, list<set<string>>, which each item in our 
// list will be tested against the decisionTree
let solve input =
    Regex.Matches(input, @"(\(((?<s>\()|[^()]|(?<-s>\)))*\)(?(s)(?!)))\s+\d+\s+((\S+\s\d(.+)?\s*)+)")
    |> Seq.cast<Match>
    |> Seq.map (fun m -> fst(parse(tokenize(m.Groups.[1].Value))), [for a in m.Groups.[3].Value.Trim().Split([|'\n'|]) -> set(a.Split([|' '|])) ])
    |> Seq.iteri (fun i (tree, testCases) ->
            printfn "Case #%i" (i+1)
            testCases |> Seq.iter (fun testCase -> printfn "%.7F" (eval (testCase, tree)))
        )

回答23:

R in 280 bytes

Note: On the standard distribution of R (as of v. 2.9.2), this program does not pass the large input and fails on just Case 28 (which is nested to 99 levels), generating a "contextstack overflow". To fix this, modify the line in src/main/gram.c that reads

#define CONTEXTSTACK_SIZE 50

and replace the 50 with something like 500. Then recompile. Et voilà!

n=0
g=gsub
eval(parse(text=g('[^
]* [0-9]+( [^
]*|
)','f=c(\\1)
cat(eval(d),"
")
',g('
\\(','
cat("Case #",n<-n+1,":
",sep="")
d=expression(',g('" "','","',g(')\\s*\\(',',',g(' *("[a-z]+")\\s*\\(','*ifelse(\\1%in%f,',g('([a-z]+)','"\\1"',paste(readLines('A'),collapse='
')))))))))

Usage (requires renaming input): cp input.in A; R -q --slave -f thisfile.R >output.out

来源：https://stackoverflow.com/questions/1433263/code-golf-decision-tree