結果

問題 No.90 品物の並び替え
ユーザー guricerin
提出日時 2019-06-10 15:19:45
言語 Rust
(1.83.0 + proconio)
結果
AC  
実行時間 25 ms / 5,000 ms
コード長 4,780 bytes
コンパイル時間 17,827 ms
コンパイル使用メモリ 378,528 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-06 00:41:04
合計ジャッジ時間 13,708 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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

// Original: https://github.com/tanakh/competitive-rs
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::BTreeMap;
fn main() {
    input!(n: usize, m: usize, items: [(usize, usize, usize); m]);
    let mut score = vec![vec![0usize; n]; n];
    for &(i1, i2, s) in items.iter() {
        score[i1][i2] = s;
    }
    let mut nums = (0..n).map(|x| x).collect::<Vec<usize>>();
    let mut ans = 0;
    loop {
        let mut tmp = 0;
        // l < r となるようにループ
        for r in 0..n {
            for l in 0..r {
                tmp += score[nums[l]][nums[r]];
            }
        }
        ans = max(ans, tmp);
        if !nums.next_permutation() {
            break;
        }
    }
    println!("{}", ans);
}
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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