結果

問題 No.90 品物の並び替え
ユーザー sino
提出日時 2021-04-01 15:02:52
言語 Rust
(1.83.0 + proconio)
結果
AC  
実行時間 348 ms / 5,000 ms
コード長 4,036 bytes
コンパイル時間 12,906 ms
コンパイル使用メモリ 404,992 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-12-17 19:05:20
合計ジャッジ時間 13,489 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 9
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#[allow(unused_macros)]
macro_rules! read {
    ([$t:ty] ; $n:expr) =>
        ((0..$n).map(|_| read!([$t])).collect::<Vec<_>>());
    ($($t:ty),+ ; $n:expr) =>
        ((0..$n).map(|_| read!($($t),+)).collect::<Vec<_>>());
    ([$t:ty]) =>
        (rl().split_whitespace().map(|w| w.parse().unwrap()).collect::<Vec<$t>>());
    ($t:ty) =>
        (rl().parse::<$t>().unwrap());
    ($($t:ty),*) => {{
        let buf = rl();
        let mut w = buf.split_whitespace();
        ($(w.next().unwrap().parse::<$t>().unwrap()),*)
    }};
}
#[allow(dead_code)]
fn rl() -> String {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    buf.trim_end().to_owned()
}
fn main() {
    let (n, m) = read!(usize, usize);
    let table = read!([usize]; m);
    let mut max_score = 0;
    // let mut best_order: Vec<usize> = (0..n).collect();
    let mut order: Vec<usize> = (0..n).collect();
    loop {
        let mut score = 0;
        // calc point.
        for data in &table {
            let (item1, item2, s) = (data[0], data[1], data[2]);
            // find index of item1
            let mut i1 = 0;
            for i in 0..n {
                if order[i] == item1 {
                    i1 = i;
                    break;
                }
            }
            for i in i1+1..n {
                if order[i] == item2 {
                    score += s;
                    break;
                }
            }
        }
        // update best order
        if score >= max_score {
            // best_order = order.clone();
            max_score = score;
        }
        // update order
        if !order.next_permutation() {
            break;
        }
    }
    // println!("{:?}", best_order);
    println!("{}", max_score);
}
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }
        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }
        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }
        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }
        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);
        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();
        true
    }
    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }
        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }
        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }
        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();
        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }
        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);
        true
    }
}
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