use std::cmp::*; // https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 macro_rules! input { ($($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)*} }; } 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)*} }; } 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, usize1) => (read_value!($next, usize) - 1); ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error")); } fn main() { // In order to avoid potential stack overflow, spawn a new thread. let stack_size = 104_857_600; // 100 MB let thd = std::thread::Builder::new().stack_size(stack_size); thd.spawn(|| solve()).unwrap().join().unwrap(); } const MOD: i64 = 1_000_000_007; fn dfs( v: usize, g: &[Vec<(usize, i64)>], k: i64, vis: &mut [bool], eq: &mut Vec, pot: &mut [(i64, i64)], p: (i64, i64), ) -> (i64, i64) { let (mut lo, mut hi) = if p.0 == 1 { (1 - p.1, k - p.1) } else { (p.1 - k, p.1 - 1) }; if vis[v] { if pot[v].0 == p.0 { return if pot[v].1 == p.1 { (lo, hi) } else { (1, 0) // contradiction }; } eq.push((p.1 - pot[v].1) * 2 / (pot[v].0 - p.0)); return (lo, hi); } vis[v] = true; pot[v] = p; for &(w, c) in &g[v] { let (sublo, subhi) = dfs(w, g, k, vis, eq, pot, (-p.0, c - p.1)); lo = max(lo, sublo); hi = min(hi, subhi); } (lo, hi) } fn calc(n: usize, xyz: &[(usize, usize, i64)], k: i64) -> i64 { let mut g = vec![vec![]; n]; for &(x, y, z) in xyz { g[x].push((y, z)); g[y].push((x, z)); } let mut vis = vec![false; n]; let mut ans = 1; let mut pot = vec![(0, 0); n]; for i in 0..n { let mut eq = vec![]; if !vis[i] { let (mi, ma) = dfs(i, &g, k, &mut vis, &mut eq, &mut pot, (1, 0)); eq.dedup(); if mi > ma || eq.len() >= 2 { return 0; } if eq.len() == 1 { let x = eq[0]; if x % 2 != 0 || mi > x / 2 || x / 2 > ma { return 0; } } else { ans = ans * (ma - mi + 1) % MOD; } } } ans } // https://yukicoder.me/problems/no/1502 (3.5) // <= k に変換すれば DFS で制約を見つけるだけになる。 fn solve() { input! { n: usize, m: usize, k: i64, xyz: [(usize1, usize1, i64); m], } println!("{}", (calc(n, &xyz, k) - calc(n, &xyz, k - 1) + MOD) % MOD); }