Ieronim,
I give. While I understand your admittedly fast code - it is extremely hard to follow. The cleanest code I could come up with using a bi-directional BFS is as fast as your transform() before following my recommendation. That is to say, it is about half as fast as your current version:

use constant SRC => 0;
use constant TGT => 1;
sub find_path2 {
    my ($src, $tgt, $list) = @_;
    my (@src_work, @tgt_work, %path);
    for my $dir (SRC, TGT) {
        my ($word, $work) = $dir == SRC ? ($src, \@src_work) : ($tgt, 
+\@tgt_work);
        $path{$word}[$dir] = -1;
        for (@{$list->{$word}}) {
            push @$work, $_;
            $path{$_}[$dir] = $word;
        }
    }
    while (1) {
        for my $dir (SRC, TGT) {
            my @next;
            my $work = $dir == SRC ? \@src_work : \@tgt_work;
            for my $word (@$work) {
                return build_path(\%path, $word) if $path{$word}[abs($
+dir - 1)];
                for (@{$list->{$word}}) {
                    next if $path{$_}[$dir];
                    push @next, $_;
                    $path{$_}[$dir] = $word;
                }
            }
            @$work = @next;
        }
        return 'Path not found' if ! @src_work && ! @tgt_work;
    }
}
sub build_path {
    my ($tree, $node) = @_;
    my $path = "-$node";
    for my $dir (SRC, TGT) {
        my $word = $tree->{$node}[$dir];
        while ($word ne '-1') {
            $path = $dir == SRC ? "-$word$path" : "$path-$word";
            $word = $tree->{$word}[$dir];
        }
    }
    return substr($path, 1);
}
[download]

Update (2006-07-03): Simply moving return build_path(\%path, $word) if $path{$word}[abs($dir - 1)]; to the inner most for loop and s/word/_/ brings the performance much closer. See the update in this node for an explanation why.

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