結果

問題 No.404 部分門松列
ユーザー koba-e964
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
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 31
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 #
プレゼンテーションモードにする

#[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 2
SegTree {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 underflow
let 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 underflow
let 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 MB
let thd = std::thread::Builder::new().stack_size(stack_size);
thd.spawn(|| solve()).unwrap().join().unwrap();
}
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