結果
問題 | No.2992 Range ABCD String Query |
ユーザー |
|
提出日時 | 2025-02-16 23:15:33 |
言語 | Rust (1.83.0 + proconio) |
結果 |
AC
|
実行時間 | 244 ms / 6,000 ms |
コード長 | 5,144 bytes |
コンパイル時間 | 23,900 ms |
コンパイル使用メモリ | 384,240 KB |
実行使用メモリ | 49,640 KB |
最終ジャッジ日時 | 2025-02-16 23:16:15 |
合計ジャッジ時間 | 37,506 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 41 |
ソースコード
#[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/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, chars) => {read_value!($next, String).chars().collect::<Vec<char>>()};($next:expr, usize1) => (read_value!($next, usize) - 1);($next:expr, [ $t:tt ]) => {{let len = read_value!($next, usize);read_value!($next, [$t; len])}};($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error"));}// Segment Tree. This data structure is useful for fast folding on intervals of an array// whose elements are elements of monoid I. Note that constructing this tree requires the identity// element of I and the operation of I.// Verified by: yukicoder No. 2220 (https://yukicoder.me/submissions/841554)struct SegTree<I, BiOp> {n: usize,orign: usize,dat: Vec<I>,op: BiOp,e: I,}impl<I, BiOp> SegTree<I, BiOp>where BiOp: Fn(I, I) -> I,I: Copy {pub fn new(n_: usize, op: BiOp, e: I) -> Self {let mut n = 1;while n < n_ { n *= 2; } // n is a power of 2SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e}}// ary[k] <- vpub fn update(&mut self, idx: usize, v: I) {debug_assert!(idx < self.orign);let mut k = idx + self.n - 1;self.dat[k] = v;while k > 0 {k = (k - 1) / 2;self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);}}// [a, b) (half-inclusive)// http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/#[allow(unused)]pub fn query(&self, rng: std::ops::Range<usize>) -> I {let (mut a, mut b) = (rng.start, rng.end);debug_assert!(a <= b);debug_assert!(b <= self.orign);let mut left = self.e;let mut right = self.e;a += self.n - 1;b += self.n - 1;while a < b {if (a & 1) == 0 {left = (self.op)(left, self.dat[a]);}if (b & 1) == 0 {right = (self.op)(self.dat[b - 1], right);}a = a / 2;b = (b - 1) / 2;}(self.op)(left, right)}}const INF: i32 = 1 << 29;type D = [[i32; 4]; 4];fn mul(a: D, b: D) -> D {let mut c = [[INF; 4]; 4];for i in 0..4 {for k in i..4 {for j in k..4 {c[i][j] = min(c[i][j], a[i][k] + b[k][j]);}}}c}// https://yukicoder.me/problems/no/2992 (3.5)// dp で解くことを考えると tropical semiring の上の行列の積になるので、セグメント木が使える。fn main() {#[allow(unused)]let out = std::io::stdout();#[allow(unused)]let mut out = BufWriter::new(out.lock());#[allow(unused)]macro_rules! puts {($($format:tt)*) => (let _ = write!(out,$($format)*););}#[allow(unused)]macro_rules! putvec {($v:expr) => {for i in 0..$v.len() {puts!("{}{}", $v[i], if i + 1 == $v.len() {"\n"} else {" "});}}}input! {n: usize, q: usize,s: chars,qs: [(i32, usize1, String); q],}let mut lets = [[[INF; 4]; 4]; 4];for i in 0..4 {for j in 0..4 {for k in j..4 {lets[i][j][k] = if i == k { 0 } else { 1 };}}}let mut zero = [[INF; 4]; 4];for i in 0..4 {zero[i][i] = 0;}let mut st = SegTree::new(n, mul, zero);for i in 0..n {let idx = (s[i] as u8 - b'A') as usize;st.update(i, lets[idx]);}for (ty, q0, q1) in qs {if ty == 1 {let x = q0;let c = q1.chars().next().unwrap();let idx = (c as u8 - b'A') as usize;st.update(x, lets[idx]);} else {let l = q0;let r = q1.parse::<usize>().unwrap();let res = st.query(l..r);let mut mi = INF;for i in 0..4 {mi = mi.min(res[0][i]);}puts!("{mi}\n");}}}