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@rossmurray
Created March 18, 2013 04:09
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Naive, slow solution for google code jam 2009 qualification problem A, "Alien Language." Rust 0.6 (incoming branch)
extern mod std;
use core::io::ReaderUtil;
use core::vec;
use core::vec::*;
fn main() {
let input = get_lines();
let first_line: &str = input[0];
let toks = str::split_char(first_line, ' ');
let ldn = map(toks, |&s| uint::from_str(s).get());
let (d, n) = (ldn[1], ldn[2]);
let dictionary = slice(input, 1, d+1);
let test_cases = slice(input, d+1, n+d+1);
for test_cases.eachi |i, &t| {
let answer = num_matching(dictionary, t);
io::println(fmt!("Case #%?: %d", i, answer));
}
}
fn num_matching(dic: &[~str], lex: &str) -> int {
let mut result = 0;
let combinations = parse_combinations(lex);
for combinations.each |c| {
if contains(dic, c) {
result += 1;
}
}
result
}
fn parse_combinations(lex: &str) -> ~[~str] {
let cvec = break_str_into_groups(lex);
let p = product_n(cvec);
p
}
fn break_str_into_groups(lex: &str) -> ~[~[~str]] {
let mut cvec = ~[];
let mut inner_vec = ~[];
let mut in_parens = false;
let pieces = map(str::chars(lex), |&c| { str::from_char(c) } );
for pieces.each |&c| {
if c == ~"(" {
in_parens = true;
}
else if c == ~")" {
in_parens = false;
let mut tmp = ~[];
for inner_vec.each |&t| { tmp.push(t); }
cvec.push(tmp);
inner_vec.clear();
}
else {
if in_parens == true {
inner_vec.push(c);
}
else {
cvec.push(~[c]);
}
}
}
cvec
}
fn product_n(v: &[~[~str]]) -> ~[~str] {
match v.len() {
n if n < 2 => fail!(),
n if n == 2 => product_pair(v[0], v[1]),
n => product_pair(v[0], product_n(slice(v, 1, n)))
}
}
fn product_pair(a: &[~str], b: &[~str]) -> ~[~str] {
let mut result = ~[];
for a.each |&i| {
for b.each |&j| {
result.push(i + j);
}
}
result
}
fn get_lines() -> ~[~str] {
let mut result = ~[];
let reader = io::stdin();
let util = @reader as @ReaderUtil;
while !reader.eof() {
result.push(util.read_line());
}
result
}
@rossmurray
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No regular expressions yet, so I just solve it the naive way using cartesian product. It's too slow to solve even the "small" input sample google provides (or runs out of memory, not sure).

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