結果

問題 No.3305 Shift Sort
ユーザー koba-e964
提出日時 2025-10-08 12:17:48
言語 Rust
(1.83.0 + proconio)
結果
AC  
実行時間 176 ms / 2,000 ms
コード長 6,788 bytes
コンパイル時間 25,891 ms
コンパイル使用メモリ 398,780 KB
実行使用メモリ 34,856 KB
最終ジャッジ日時 2025-10-08 12:18:21
合計ジャッジ時間 32,495 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
other AC * 20
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 # 

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
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。
fn main() {
    #[allow(unused)]
    let out = std::io::stdout();
    #[allow(unused)]
    let mut out = BufWriter::new(out.lock());
    #[allow(unused)]
    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]);
    }
}
0