#1126857 (Rust) No.3305 Shift Sort

• 問題: No.3305 Shift Sort
• ユーザー: koba-e964
• 提出日時: 2025-10-08 12:18:58
• 言語: Rust (1.83.0 + proconio)
• 結果: AC
• 実行時間: 170 ms / 2,000 ms
• コード長: 6,644 bytes
• コンパイル時間: 24,413 ms
• コンパイル使用メモリ: 401,488 KB
• 実行使用メモリ: 34,888 KB
• 最終ジャッジ日時: 2025-10-08 12:19:29
• 合計ジャッジ時間: 17,975 ms
• ジャッジサーバーID （参考情報）: judge5 / judge4

ファイルパターン	結果
other	AC * 20

ソースコード

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, $t:ty) => ($next().parse::<$t>().expect("Parse error"));
}

// Segment Tree. This data structure is useful for fast folding on intervals of an array
// whose elements are elements of monoid I. Note that constructing this tree requires the identity
// element of I and the operation of I.
// Verified by: yukicoder No. 2220 (https://yukicoder.me/submissions/841554)
struct SegTree<I, BiOp> {
    n: usize,
    orign: usize,
    dat: Vec<I>,
    op: BiOp,
    e: I,
}

impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    pub fn new(n_: usize, op: BiOp, e: I) -> Self {
        let mut n = 1;
        while n < n_ { n *= 2; } // n is a power of 2
        SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e}
    }
    // ary[k] <- v
    pub fn update(&mut self, idx: usize, v: I) {
        debug_assert!(idx < self.orign);
        let mut k = idx + self.n - 1;
        self.dat[k] = v;
        while k > 0 {
            k = (k - 1) / 2;
            self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
        }
    }
    // [a, b) (half-inclusive)
    // http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/
    #[allow(unused)]
    pub fn query(&self, rng: std::ops::Range<usize>) -> I {
        let (mut a, mut b) = (rng.start, rng.end);
        debug_assert!(a <= b);
        debug_assert!(b <= self.orign);
        let mut left = self.e;
        let mut right = self.e;
        a += self.n - 1;
        b += self.n - 1;
        while a < b {
            if (a & 1) == 0 {
                left = (self.op)(left, self.dat[a]);
            }
            if (b & 1) == 0 {
                right = (self.op)(self.dat[b - 1], right);
            }
            a = a / 2;
            b = (b - 1) / 2;
        }
        (self.op)(left, right)
    }
}

// Depends on: datastr/SegTree.rs
// Verified by: yukicoder No. 2220 (https://yukicoder.me/submissions/841554)
impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    // Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp
    #[allow(unused)]
    fn max_right<F: Fn(I) -> bool>(
        &self, rng: std::ops::RangeFrom<usize>, f: &F,
    ) -> usize {
        let mut l = rng.start;
        assert!(f(self.e));
        if l == self.orign {
            return self.orign;
        }
        l += self.n - 1;
        let mut sm = self.e;
        loop {
            while l % 2 == 1 {
                l = (l - 1) / 2;
            }
            if !f((self.op)(sm, self.dat[l])) {
                while l < self.n - 1 {
                    l = 2 * l + 1;
                    let val = (self.op)(sm, self.dat[l]);
                    if f(val) {
                        sm = val;
                        l += 1;
                    }
                }
                return std::cmp::min(self.orign, l + 1 - self.n);
            }
            sm = (self.op)(sm, self.dat[l]);
            l += 1;
            if (l + 1).is_power_of_two() { break; }
        }
        self.orign
    }
    // Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp
    #[allow(unused)]
    fn min_left<F: Fn(I) -> bool>(
        &self, rng: std::ops::RangeTo<usize>, f: &F,
    ) -> usize {
        let mut r = rng.end;
        if !f(self.e) {
            return r + 1;
        }
        if r == 0 {
            return 0;
        }
        r += self.n - 1;
        let mut sm = self.e;
        loop {
            r -= 1;
            while r > 0 && r % 2 == 0 {
                r = (r - 1) / 2;
            }
            if !f((self.op)(self.dat[r], sm)) {
                while r < self.n - 1 {
                    r = 2 * r + 2;
                    let val = (self.op)(self.dat[r], sm);
                    if f(val) {
                        sm = val;
                        r -= 1;
                    }
                }
                return r + 2 - self.n;
            }
            sm = (self.op)(self.dat[r], sm);
            if (r + 1).is_power_of_two() { break; }
        }
        0
    }
}

// https://yukicoder.me/problems/no/3305 (3)
// 問題を言い換えると、 a[l, r) で自分の左側に自分より大きい要素がある要素は何個かという問題になる。
// a の中で a[j] > a[i] なる最大の j < i を i ごとに計算しておき、
// query 先読みで左端 l が現れる時にセグメント木の該当する要素に 1 を足せばよい。
// Tags: query-look-ahead, rotation
fn main() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {($($format:tt)*) => (let _ = write!(out,$($format)*););}
    input! {
        n: usize, q: usize,
        a: [usize; n],
        lr: [(usize1, usize); q],
    }
    let mut ma_left = vec![0; n]; // offset: -1
    {
        let mut st = SegTree::new(n, |x, y| x.max(y), -1);
        for i in 0..n {
            let l = st.min_left(..i, &|x| x <= a[i] as i32);
            ma_left[i] = l;
            st.update(i, a[i] as i32);
        }
    }
    let mut qs = vec![vec![]; n];
    for i in 0..q {
        let (l, r) = lr[i];
        qs[l].push((i, r));
    }
    let mut added = vec![vec![]; n];
    for i in 0..n {
        added[ma_left[i]].push(i);
    }
    let mut ans = vec![0; q];
    let mut st = SegTree::new(n, |x, y| x + y, 0);
    for l in (0..n).rev() {
        for &(i, r) in &qs[l] {
            ans[i] = st.query(l..r);
        }
        for &i in &added[l] {
            st.update(i, 1);
        }
    }
    for i in 0..q {
        puts!("{}\n", ans[i]);
    }
}