// 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, $t)).collect::>() }; ($next:expr, chars) => { read_value!($next, String).chars().collect::>() }; ($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::>(); 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 LexicalPermutation for [T] where T: Ord, { /// Original author in Rust: Thomas Backman 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 } }