#721142 (Rust) No.1442 I-wate Shortest Path Problem

• 問題: No.1442 I-wate Shortest Path Problem
• ユーザー: koba-e964
• 提出日時: 2021-12-01 23:37:03
• 言語: Rust (1.77.0 + proconio)
• 結果: AC
• 実行時間: 768 ms / 3,000 ms
• コード長: 6,715 bytes
• コンパイル時間: 14,089 ms
• コンパイル使用メモリ: 387,600 KB
• 実行使用メモリ: 57,848 KB
• 最終ジャッジ日時: 2024-07-05 01:38:57
• 合計ジャッジ時間: 24,989 ms
• ジャッジサーバーID （参考情報）: judge5 / judge2

入力	結果	実行時間; 実行使用メモリ
testcase_00	AC	1 ms; 5,248 KB
testcase_01	AC	1 ms; 5,376 KB
testcase_02	AC	7 ms; 5,376 KB
testcase_03	AC	40 ms; 5,376 KB
testcase_04	AC	7 ms; 5,376 KB
testcase_05	AC	5 ms; 5,376 KB
testcase_06	AC	39 ms; 5,376 KB
testcase_07	AC	4 ms; 5,376 KB
testcase_08	AC	32 ms; 5,376 KB
testcase_09	AC	12 ms; 5,376 KB
testcase_10	AC	45 ms; 5,376 KB
testcase_11	AC	39 ms; 5,376 KB
testcase_12	AC	452 ms; 47,500 KB
testcase_13	AC	235 ms; 42,752 KB
testcase_14	AC	353 ms; 44,388 KB
testcase_15	AC	345 ms; 45,552 KB
testcase_16	AC	477 ms; 47,876 KB
testcase_17	AC	757 ms; 52,108 KB
testcase_18	AC	762 ms; 52,136 KB
testcase_19	AC	519 ms; 48,680 KB
testcase_20	AC	764 ms; 52,036 KB
testcase_21	AC	768 ms; 52,080 KB
testcase_22	AC	300 ms; 52,736 KB
testcase_23	AC	536 ms; 57,848 KB
testcase_24	AC	212 ms; 45,604 KB
testcase_25	AC	500 ms; 50,240 KB
testcase_26	AC	214 ms; 46,136 KB

ソースコード

use std::cmp::*;
use std::collections::*;
use std::io::{Write, BufWriter};
// 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, $t)).collect::<Vec<_>>()
    };
    ($next:expr, usize1) => (read_value!($next, usize) - 1);
    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        read_value!($next, [$t; len])
    }};
    ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error"));
}

// Lowest Common Ancestor. Call lca(x, y) to get the lca of them.
// Many-rooted version.
// Verified by: https://yukicoder.me/submissions/714482
// This library uses O(n) stack space. 
pub struct LCA {
    n: usize,
    bn: usize,
    parent: Vec<usize>, // r is root <=> parent[r] = r
    dep: Vec<usize>,
    lca_tbl: Vec<Vec<usize>>
}

impl LCA {
    fn dfs(&mut self, edges: &[Vec<usize>], v: usize, par: usize, d: usize) {
        self.parent[v] = par;
        self.dep[v] = d;
        for &u in edges[v].iter() {
            if u != par {
                self.dfs(edges, u, v, d + 1);
            }
        }
    }
    fn lca_init(&mut self) {
        let n = self.n;
        for v in 0 .. n {
            self.lca_tbl[v] = vec![0; self.bn + 1];
            self.lca_tbl[v][0] = self.parent[v];
        }
        for i in 1..self.bn + 1 {
            for v in 0..n {
                self.lca_tbl[v][i] =
                    self.lca_tbl[self.lca_tbl[v][i - 1]][i - 1];
            }
        }
    }
    pub fn lca(&self, mut x: usize, mut y: usize) -> usize {
        let dx = self.dep[x];
        let mut dy = self.dep[y];
        if dx > dy {
            return self.lca(y, x);
        }
        for l in (0..self.bn + 1).rev() {
            if dy - dx >= 1 << l {
                y = self.lca_tbl[y][l];
                dy -= 1 << l;
            }
        }
        assert_eq!(dx, dy);
        
        if x == y {
            return x;
        }
        for l in (0..self.bn + 1).rev() {
            if self.lca_tbl[x][l] != self.lca_tbl[y][l] {
	        x = self.lca_tbl[x][l];
	        y = self.lca_tbl[y][l];
            }
        }
        self.lca_tbl[x][0]
    }
    pub fn new(edges: &[Vec<usize>], roots: &[usize]) -> Self {
        let n = edges.len();
        let bn = (n.next_power_of_two() - 1).count_ones() as usize;
        let mut ret = LCA {
            n: n, bn: bn, parent: vec![0; n], dep: vec![0; n],
            lca_tbl: vec![Vec::new(); n] };
        for &r in roots {
            ret.dfs(edges, r, r, 0);
        }
        ret.lca_init();
        ret
    }
}

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();
}

fn dfs(v: usize, par: usize, g: &[Vec<(usize, i64)>], dep: &mut [i64], x: i64) {
    dep[v] = x;
    for &(w, c) in &g[v] {
        if w == par { continue; }
        dfs(w, v, g, dep, x + c);
    }
}

// https://yukicoder.me/problems/no/1442 (3)
// 航空会社の端点から別の航空会社の端点への距離を K^2 通りすべて計算しておく。また航空会社の端点から各頂点への距離を求めておく。クエリの答えは、U, V 用の頂点および航空会社用の頂点の合計 K+2 頂点のグラフにおける U-V 間の最短距離である。航空会社用の頂点の部分については最短距離をあらかじめ求めておくことができる。
// 計算量は O(NK log N + Q (log N + K^2) + K^3) である。
fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {($($format:tt)*) => (let _ = write!(out,$($format)*););}
    input! {
        n: usize, k: usize,
        abc: [(usize1, usize1, i64); n - 1],
        mpx_tmp: [([i64], i64); k],
        q: usize,
        uv: [(usize1, usize1); q],
    }
    let mut g = vec![vec![]; n];
    let mut g_uw = vec![vec![]; n];
    for &(a, b, c) in &abc {
        g[a].push((b, c));
        g[b].push((a, c));
        g_uw[a].push(b);
        g_uw[b].push(a);
    }
    let mut px = vec![];
    for i in 0..k {
        let (mut tmp, rest) = mpx_tmp[i].clone();
        tmp.push(rest);
        let p = tmp.remove(0);
        let mut x = Vec::with_capacity(tmp.len());
        for &tmp in &tmp {
            x.push(tmp as usize - 1);
        }
        px.push((p, x));
    }
    const INF: i64 = 1 << 50;
    let mut pd = vec![vec![0; k]; k];
    let mut from = vec![vec![]; k];
    for i in 0..k {
        let mut dist = vec![INF; n];
        let (p1, ref x) = px[i];
        let mut que = BinaryHeap::new();
        for &x in x {
            que.push((Reverse(0), x));
        }
        while let Some((Reverse(d), v)) = que.pop() {
            if dist[v] <= d { continue; }
            dist[v] = d;
            for &(w, c) in &g[v] {
                que.push((Reverse(d + c), w));
            }
        }
        for j in 0..n {
            dist[j] = 2 * dist[j] + p1;
        }
        for j in 0..k {
            if i == j { continue; }
            let (p2, ref x) = px[j];
            let mut mi = INF;
            for &x in x {
                mi = min(mi, dist[x]);
            }
            pd[i][j] = p2 + mi;
        }
        from[i] = dist;
    }
    // Floyd-Warshall
    for l in 0..k {
        for i in 0..k {
            for j in 0..k {
                pd[i][j] = min(pd[i][j], pd[i][l] + pd[l][j]);
            }
        }
    }
    let lca = LCA::new(&g_uw, &[0]);
    let mut dep = vec![0; n];
    dfs(0, n, &g, &mut dep, 0);
    for &(u, v) in &uv {
        let l = lca.lca(u, v);
        let mut mi = 2 * (dep[u] + dep[v] - 2 * dep[l]);
        for i in 0..k {
            for j in 0..k {
                mi = min(mi, pd[i][j] + from[i][u] + from[j][v]);
            }
        }
        puts!("{}\n", mi / 2);
    }
}