結果
問題 | No.404 部分門松列 |
ユーザー |
|
提出日時 | 2017-03-04 03:08:33 |
言語 | Rust (1.83.0 + proconio) |
結果 |
AC
|
実行時間 | 1,976 ms / 2,000 ms |
コード長 | 6,006 bytes |
コンパイル時間 | 14,706 ms |
コンパイル使用メモリ | 379,756 KB |
実行使用メモリ | 23,644 KB |
最終ジャッジ日時 | 2024-09-25 18:00:47 |
合計ジャッジ時間 | 39,061 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge4 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 31 |
ソースコード
#[allow(unused_imports)]use std::cmp::*;#[allow(unused_imports)]use std::collections::*;use std::io::Read;#[allow(dead_code)]fn getline() -> String {let mut ret = String::new();std::io::stdin().read_line(&mut ret).ok().unwrap();ret}fn get_word() -> String {let mut stdin = std::io::stdin();let mut u8b: [u8; 1] = [0];loop {let mut buf: Vec<u8> = Vec::with_capacity(16);loop {let res = stdin.read(&mut u8b);if res.unwrap_or(0) == 0 || u8b[0] <= b' ' {break;} else {buf.push(u8b[0]);}}if buf.len() >= 1 {let ret = String::from_utf8(buf).unwrap();return ret;}}}#[allow(dead_code)]fn get<T: std::str::FromStr>() -> T { get_word().parse().ok().unwrap() }/*** Segment Tree. This data structure is useful for fast folding on intervals of an array* whose elements are elements of monoid M. Note that constructing this tree requires the identity* element of M and the operation of M.* Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581)*/struct SegTree<I, BiOp> {n: 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 2SegTree {n: n, dat: vec![e; 2 * n - 1], op: op, e: e}}/* ary[k] <- v */pub fn update(&mut self, idx: usize, v: I) {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]);}}/* l,r are for simplicity */fn query_sub(&self, a: usize, b: usize, k: usize, l: usize, r: usize) -> I {// [a,b) and [l,r) intersects?if r <= a || b <= l { return self.e; }if a <= l && r <= b { return self.dat[k]; }let vl = self.query_sub(a, b, 2 * k + 1, l, (l + r) / 2);let vr = self.query_sub(a, b, 2 * k + 2, (l + r) / 2, r);(self.op)(vl, vr)}/* [a, b] (note: inclusive) */pub fn query(&self, a: usize, b: usize) -> I {self.query_sub(a, b + 1, 0, 0, self.n)}}/// Coordinate compression/// Returns a vector of usize, with i-th element the "rank" of a[i] in a./// The property forall i. inv_map[ret[i]] == a[i] holds.fn coord_compress<T: Ord + std::fmt::Debug>(a: &[T])-> (Vec<usize>, Vec<&T>) {let n = a.len();let mut cp: Vec<(&T, usize)> = (0 .. n).map(|i| (&a[i], i)).collect();cp.sort();let mut inv_map = Vec::new();let mut prev: Option<&T> = None;let mut ret = vec![0; n];let mut cnt = 0;for (v, i) in cp {if prev == Some(v) {ret[i] = cnt - 1;continue;}ret[i] = cnt;inv_map.push(v);prev = Some(v);cnt += 1;}for i in 0 .. n {assert_eq!(*inv_map[ret[i]], a[i]);}(ret, inv_map)}fn calc_three_steps(a: &[usize]) -> Vec<i64> {let n = a.len();// Shifted by 1 (right) to avoid subtraction underflowlet mut st = SegTree::new(n + 1, |x, y| x + y, 0);let mut st_sq = SegTree::new(n + 1, |x, y| x + y, 0);let mut ret = vec![0; n];for i in 0 .. n {let tmp = st.query(a[i] + 1, a[i] + 1) + 1;let stsum = st.query(1, a[i]);ret[i] = (stsum * stsum - st_sq.query(1, a[i])) / 2;st.update(a[i] + 1, tmp);st_sq.update(a[i] + 1, tmp * tmp);}ret}fn calc_three_any(a: &[usize]) -> Vec<i64> {let n = a.len();// Shifted by 1 (right) to avoid subtraction underflowlet mut st = SegTree::new(n + 1, |x, y| x + y, 0);let mut st_sq = SegTree::new(n + 1, |x, y| x + y, 0);let mut ret = vec![0; n];for i in 0 .. n {let tmp = st.query(a[i] + 1, a[i] + 1) + 1;st.update(a[i] + 1, tmp);st_sq.update(a[i] + 1, tmp * tmp);}for i in 0 .. n {let stsum = st.query(1, a[i]);ret[i] = (stsum * stsum - st_sq.query(1, a[i])) / 2;}ret}// Finds #{(j, k) | j < i < k, a[j] < a[i] > a[k], a[j] != a[k]} for every i.fn calc_max_aux(a: &[i64]) -> Vec<i64> {let n = a.len();let (mut a, _) = coord_compress(a);let three = calc_three_steps(&a);let mut ret = calc_three_any(&a);a.reverse();let three_rev = calc_three_steps(&a);for i in 0 .. n {ret[i] += -three[i] - three_rev[n - 1 - i];}ret}// Precomputation// Counts how many kadomatsu seqs. with the center a[i] can be made.// O(n * log(n))fn calc_aux(a: &[i64]) -> Vec<i64> {let n = a.len();let mut aux: Vec<i64> = calc_max_aux(a);let mut a = a.to_vec();for v in a.iter_mut() {*v *= -1;}let res2 = calc_max_aux(&a);for i in 0 .. n {aux[i] += res2[i];}aux}fn solve() {let n = get();let a: Vec<i64> = (0 .. n).map(|_| get()).collect();let aux = calc_aux(&a);const INF: i64 = 1 << 60;let mut acc = vec![(0, 0); n];for i in 0 .. n {acc[i] = (a[i], aux[i]);}acc.push((-INF, 0));acc.sort();for i in 0 .. n + 1 {acc[i].1 += if i == 0 { 0 } else { acc[i - 1].1 };}let q = get();for _ in 0 .. q {let l: i64 = get();let h: i64 = get();let upper = acc.binary_search(&(h, INF)).unwrap_err();let lower = acc.binary_search(&(l, -INF)).unwrap_err();println!("{}", acc[upper - 1].1 - acc[lower - 1].1);}}fn main() {// In order to avoid potential stack overflow, spawn a new thread.let stack_size = 104_857_600; // 100 MBlet thd = std::thread::Builder::new().stack_size(stack_size);thd.spawn(|| solve()).unwrap().join().unwrap();}