CodeGolf:找到独特的路径

时间:2010-12-27 09:10:02

标签: language-agnostic graph code-golf rosetta-stone

这是一个非常简单的想法,在这个pastebin我发布了一些数字。这些代表有向图的节点。 stdin的输入将是格式,(它们将是数字,我将在此处使用示例)

c d
q r
a b
b c
d e
p q

所以x y表示xy相关联(反之亦然)

该示例中有2条路径。 a->b->c->d->ep->q->r

您需要打印该图表中的所有唯一路径 输出的格式应为

a->b->c->d->e
p->q->r

备注

  1. 您可以假设选择的数字使得一条路径不与另一条路径相交(一个节点属于一条路径)
  2. 这些对是随机顺序的。
  3. 它们不止一条路径,它们可以有不同的长度。
  4. 所有数字均小于1000.
  5. 如果您需要更多详细信息,请发表评论。我会根据需要修改。

    无耻插头

    对于那些喜欢Codegolf的人,请在Area51为其自己的网站提交:)(对于那些不喜欢它的人,也请支持它,所以我们将不会让您失望。 ..)

9 个答案:

答案 0 :(得分:6)

Ruby - 132 125 87

h=Hash[a=[*$<].map(&:split)]
1000.times{a.map!{|i|i+[h[i[-1]]]-[nil]}}
puts a.sort_by{|i|-i.size}.uniq(&:last).map{|i|i*'->'}

Took Nas Banov关于h.keys-h.values

的想法
h=Hash[[*$<].map &:split]
puts (h.keys-h.values).map{|i|s=i
s+='->'+i=h[i]while h[i];s}

答案 1 :(得分:5)

虽然不是答案,但以下Linux脚本绘制了输入文件的图形:

cat FILE | (echo "digraph G {"; awk '{print "\t" $1 "-> " $2;}'; echo "}") \
   | dot -Tpng > out.png && eog out.png

您需要为dot命令安装GraphvizeogGNOME的图片查看器。

在输入文件上运行,图形如下所示:

alt text

旋转90°并按比例缩小(see original

如您所见,输入图只是单链表的集合,没有共享节点且没有循环。

答案 2 :(得分:5)

C89 - 212 204个字符

#define M 1001
int t[M],r[M],a,b;main(){while(scanf("%d%d",&a,&b)>0)t[a+1]=r[a+1]=b+1;
for(a=1;a<M;a++)r[t[a]]=0;for(a=1;a<M;a++)if(r[a]){printf("%d",a-1);
for(b=t[a];b;b=t[b])printf("->%d",b-1);puts("");}}

不计算不必要的换行符。

命令:

$ wget -O - http://pastebin.com/download.php?i=R2PDGb2w | ./unique-paths

输出:

477->4->470->350->401->195->258->942->263->90->716->514->110->859->976->104->119->592->968->833->731->489->364->847->727
784->955->381->231->76->644->380->861->522->775->565->773->188->531->219->755->247->92->723->726->606
821->238->745->504->99->368->412->142->921->468->315->193->674->793->673->405->185->257->21->212->783->481->269

漂亮版:

#include <stdio.h>

int main(void)
{
    /* Note: {0} initializes all items to zero. */
    int target[1001] = {0}; /* If a → b, then target[a+1] == b+1. */
    int root[1001]   = {0}; /* If a is a root, then root[a+1] != 0. */
    int a, b, i, next;

    /* Read input numbers, setting the target of each node.
       Also, mark each source node as a root. */
    while (scanf("%d %d", &a, &b) == 2)
        target[a+1] = root[a+1] = b+1;

    /* Mark each node that is pointed to as not a root. */
    for (i = 1; i <= 1000; i++)
        root[target[i]] = 0;

    /* For each root node, print its chain. */
    for (i = 1; i <= 1000; i++) {
        if (root[i] != 0) {
            printf("%d", i-1);
            for (next = target[i]; next != 0; next = target[next])
                printf("->%d", next-1);
            printf("\n");
        }
    }

    return 0;
}

答案 3 :(得分:5)

的Python

120个字符

我喜欢它在Python中读取的轻松程度:

import sys
d=dict(map(str.split,sys.stdin))
for e in set(d)-set(d.values()):
    while e in d:print e,'->',;e=d[e]
    print e

跑过面食箱样品的结果:

784 -> 955 -> 381 -> 231 -> 76 -> 644 -> 380 -> 861 -> 522 -> 775 -> 565 -> 773 -> 188 -> 531 -> 219 -> 755 -> 247 -> 92 -> 723 -> 726 -> 606
821 -> 238 -> 745 -> 504 -> 99 -> 368 -> 412 -> 142 -> 921 -> 468 -> 315 -> 193 -> 674 -> 793 -> 673 -> 405 -> 185 -> 257 -> 21 -> 212 -> 783 -> 481 -> 269
477 -> 4 -> 470 -> 350 -> 401 -> 195 -> 258 -> 942 -> 263 -> 90 -> 716 -> 514 -> 110 -> 859 -> 976 -> 104 -> 119 -> 592 -> 968 -> 833 -> 731 -> 489 -> 364 -> 847 -> 727

答案 4 :(得分:4)

Lua,166字节

噢,最终是一个代号高手,Lua并不傻逼。额外的好东西:适用于空间分隔的任何东西(任何大小的数字,字符串......)

非高尔夫球版:

-- Read in a file from stdout filled with pairs of numbers representing nodes of a (single-)directed graph.
-- x y means x->y (but not y->x)
g={}t={}w=io.write
i=io.read"*a" -- read in numbers from stdin
for x,y in i:gmatch"(%w+) (%w+)" do -- parse pairs 
    t[y]=1 -- add y to destinations (which never can be a starting point)
    g[x]=y
end
for k,v in pairs(g) do -- go through all links
    if not t[k] then   -- only start on starting points         
        w(k)           -- write the startingpoint
        while v do     -- as long as there is a destination ...
            w('->',v)  -- write link
            v=g[v]     -- next destination
        end
        w'\n'
    end
end

高尔夫版本:

g={}t={}w=io.write for x,y in io.read"*a":gmatch"(%w+) (%w+)"do t[y]=1 g[x]=y end for k,v in pairs(g)do if not t[k]then w(k)while v do w('->',v)v=g[v]end w'\n'end end

答案 5 :(得分:3)

Haskell - 174 142 137 133个字符

import List
a#m=maybe[](\x->"->"++x++x#m)$lookup a m
q[f,s]=f\\s>>=(\a->a++a#zip f s++"\n")
main=interact$q.transpose.map words.lines

Ungolfed:

import Data.List

type Node = String

follow :: Node -> [(Node,Node)] -> String
follow node pairs = maybe "" step $ lookup node pairs
    where step next = "->" ++ next ++ follow next pairs

chains :: [[Node]] -> String
chains [firsts,seconds] = concatMap chain $ firsts \\ seconds
    where chain node = node ++ follow node pairs ++ "\n"
          pairs = zip firsts seconds

process :: [String] -> String
process = chains . transpose . map words

main :: IO ()
main=interact $ process . lines

比以前更优雅的方法,但更短!灵感来自Nas Banov关于h.keys-h.values

的想法

答案 6 :(得分:1)

PHP - 155

foreach(file($argv[1])as$x){$x=explode(' ',$x);$g[$x[0]+0]=$x[1]+0;}
foreach($g as$a=>$b)if(!in_array($a,$g)){echo$a;while($b=$g[$b])echo"->$b";echo"\n";}

整个文件看起来像:

<?php
error_reporting(0);
foreach(file($argv[1])as$x){$x=explode(' ',$x);$g[$x[0]+0]=$x[1]+0;}
foreach($g as$a=>$b)if(!in_array($a,$g)){echo$a;while($b=$g[$b])echo"->$b";echo"\n";}

运行:

$ php graph.php graph.txt

漂亮版:

$lines = file($argv[1]);
foreach ($lines as $line) {
    $vertexes = explode(' ',$line);
    $graph[$vertexes[0]+0] = $vertexes[1]+0; // the +0 forces it to an integer
}
foreach ($graph as $a => $b) {
    //searches the vertexes that are pointed to for $a
    if (!in_array($a,$graph)) {
        echo $a;
        for ($next = $b; isset($graph[$next]); $next = $graph[$next]) {
            echo "->$next";
        }
        //because the loop doesn't run one last time, like in the golfed version
        echo "->$next\n";
    }
}

答案 7 :(得分:0)

Java

372 337 304字节

更新

  1. 删除了泛型。显然,上课可以做而不是“公共”(Thnx Sean)
  2. 删除了Class,由Enum取代。 (见评论,Thnx Nabb)
  3. import java.util.*;enum M{M;{Scanner s=new Scanner(System.in);final Map g=new HashMap();while(s.hasNext()){g.put(s.nextInt(),s.nextInt());}for(int a:new HashSet<Integer>(g.keySet()){{removeAll(g.values());}}){while(g.containsKey(a)){System.out.print(a+"->");a=(Integer)g.get(a);}System.out.println(a);}}}
    

答案 8 :(得分:0)

ocaml的

402个字符

基本上是对Haskell版本的改编,其长度令我惊讶。使用Str和/或revised syntax确实可以做得更好。

open List;;open String;; let q(a,b,p)=print_string(p^b^"\n")in let rec f(a,b,p)=function []->[a,b,p]|(x,y,q)::l when x=b->f(a,y,p^q)l|(x,y,q)::l when y=a->f(x,b,q^p)l|h::t->h::(f(a,b,p)t)in let t s=let i=index s ' 'in let h=sub s 0 i in h,sub s (i+1) ((length s) -i-1),h^"->"in let s=ref []in try while true do let l=read_line ()in s:=l::!s done with End_of_file->List.iter q(fold_right f(map t !s)[])

Ungolfed:

open List;;
open String;;
let print (a,b,p) = print_string (p^b^"\n") in
let rec compose (a,b,p) = function
   [] -> [a,b,p]
  |(x,y,q)::l when x=b->compose (a,y,p^q) l
  |(x,y,q)::l when y=a->compose (x,b,q^p) l
  |h::t->h::(compose(a,b,p) t) in
let tokenize s = let i = index s ' ' in
          let h =  sub s 0 i in
          h,sub s (i+1) ((length s) -i-1),h^"->" in
let lines = ref [] in
try
  while true do
    let l = read_line () in
    lines := l::!lines
  done
with
    End_of_file-> List.iter print (fold_right compose (map tokenize !lines) [])