結果
問題 | No.2333 Slime Structure |
ユーザー |
|
提出日時 | 2023-05-31 12:35:22 |
言語 | Rust (1.83.0 + proconio) |
結果 |
AC
|
実行時間 | 237 ms / 3,000 ms |
コード長 | 5,115 bytes |
コンパイル時間 | 20,429 ms |
コンパイル使用メモリ | 379,512 KB |
実行使用メモリ | 41,016 KB |
最終ジャッジ日時 | 2024-12-28 13:42:54 |
合計ジャッジ時間 | 25,585 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 31 |
ソースコード
#[allow(unused_imports)]use std::cmp::*;#[allow(unused_imports)]use std::collections::*;use std::io::{Write, BufWriter};// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8macro_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::<$t>().expect("Parse error"));}// Segment Tree. This data structure is useful for fast folding on intervals of an array// whose elements are elements of monoid I. Note that constructing this tree requires the identity// element of I and the operation of I.// Verified by: yukicoder No. 2220 (https://yukicoder.me/submissions/841554)struct SegTree<I, BiOp> {n: usize,orign: usize,dat: Vec<I>,op: BiOp,e: I,}impl<I, BiOp> SegTree<I, BiOp>where BiOp: Fn(I, I) -> I,I: Copy {pub fn new(n_: usize, op: BiOp, e: I) -> Self {let mut n = 1;while n < n_ { n *= 2; } // n is a power of 2SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e}}// ary[k] <- vpub fn update(&mut self, idx: usize, v: I) {debug_assert!(idx < self.orign);let mut k = idx + self.n - 1;self.dat[k] = v;while k > 0 {k = (k - 1) / 2;self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);}}// [a, b) (half-inclusive)// http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/#[allow(unused)]pub fn query(&self, rng: std::ops::Range<usize>) -> I {let (mut a, mut b) = (rng.start, rng.end);debug_assert!(a <= b);debug_assert!(b <= self.orign);let mut left = self.e;let mut right = self.e;a += self.n - 1;b += self.n - 1;while a < b {if (a & 1) == 0 {left = (self.op)(left, self.dat[a]);}if (b & 1) == 0 {right = (self.op)(self.dat[b - 1], right);}a = a / 2;b = (b - 1) / 2;}(self.op)(left, right)}}const INF: i64 = 1 << 60;fn mul((ama, ami, amadiff, adiff): (i64, i64, i64, i64),(bma, bmi, bmadiff, bdiff): (i64, i64, i64, i64),) -> (i64, i64, i64, i64) {let mut madiff = max(amadiff, max(bmadiff, adiff + bma - ami));if ama > -INF {madiff = max(madiff, adiff + bma);}(max(ama, adiff + bma),min(ami, adiff + bmi),madiff,adiff + bdiff,)}// https://yukicoder.me/problems/no/2333 (3.5)// 個数が 10^5 以下であれば、セグメント木でできる。// (区間に付随する値は (max, min, 実現できる最大 diff, 区間そのものの diff) である。)// 今回の場合は座標圧縮すればよい。fn main() {let out = std::io::stdout();let mut out = BufWriter::new(out.lock());macro_rules! puts {($($format:tt)*) => (let _ = write!(out,$($format)*););}input! {n: usize,ab: [(i64, i64); n],q: usize,qs: [(i32, i64, i64); q],}let mut coo = vec![];let mut pos = 0;for &(_, b) in &ab {coo.push(pos);pos += b;}coo.push(pos);for &(ty, l, r) in &qs {coo.push(l - 1);if ty == 2 {coo.push(r);} else {coo.push(l);}}coo.sort(); coo.dedup();let m = coo.len();let mut st = SegTree::new(m, mul, (-INF, INF, -INF, 0));let mut pos = 0;for &(a, b) in &ab {let lo = coo.binary_search(&pos).unwrap();let hi = coo.binary_search(&(pos + b)).unwrap();for i in lo..hi {let t = a * (coo[i + 1] - coo[i]);st.update(i, (max(a, t), min(a, t), max(a, t), t));}pos += b;}for &(ty, l, r) in &qs {let l = l - 1;if ty == 1 {let idx = coo.binary_search(&l).unwrap();assert_eq!(coo[idx + 1], l + 1);st.update(idx, (r, r, r, r));} else {let lo = coo.binary_search(&l).unwrap();let hi = coo.binary_search(&r).unwrap();let val = st.query(lo..hi);puts!("{}\n", val.2);}}}