結果
問題 | No.2161 Black Market |
ユーザー |
|
提出日時 | 2023-07-05 02:13:47 |
言語 | Rust (1.83.0 + proconio) |
結果 |
AC
|
実行時間 | 423 ms / 7,000 ms |
コード長 | 5,283 bytes |
コンパイル時間 | 13,893 ms |
コンパイル使用メモリ | 379,744 KB |
実行使用メモリ | 13,492 KB |
最終ジャッジ日時 | 2024-07-18 20:36:25 |
合計ジャッジ時間 | 15,997 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 37 |
ソースコード
// 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, $t:ty) => ($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)}}trait Bisect<T> {fn lower_bound(&self, val: &T) -> usize;fn upper_bound(&self, val: &T) -> usize;}impl<T: Ord> Bisect<T> for [T] {fn lower_bound(&self, val: &T) -> usize {let mut pass = self.len() + 1;let mut fail = 0;while pass - fail > 1 {let mid = (pass + fail) / 2;if &self[mid - 1] >= val {pass = mid;} else {fail = mid;}}pass - 1}fn upper_bound(&self, val: &T) -> usize {let mut pass = self.len() + 1;let mut fail = 0;while pass - fail > 1 {let mid = (pass + fail) / 2;if &self[mid - 1] > val {pass = mid;} else {fail = mid;}}pass - 1}}fn calc(ab: &[(i64, i64)]) -> Vec<Vec<(i64, i64)>> {let n = ab.len();let mut ans = vec![vec![]; n + 1];for bits in 0usize..1 << n {let mut x = 0;let mut y = 0;for i in 0..n {if (bits & 1 << i) != 0 {x += ab[i].0;y += ab[i].1;}}let bc = bits.count_ones() as usize;ans[bc].push((x, y));}ans}// https://yukicoder.me/problems/no/2161 (3)// 半分全列挙 + 前半 x 個、後半 y 個 (x + y <= k) であるものを平面走査で数えるfn main() {input! {n: usize, k: usize, l: i64, p: i64,ab: [(i64, i64); n],}let fst = calc(&ab[..n / 2]);let snd = calc(&ab[n / 2..]);let mut ans = 0i64;for i in 0..k + 1 {if k - i > n - n / 2 {continue;}let mut coo = vec![];for &(_, y) in &snd[k - i] {coo.push(y);}coo.sort(); coo.dedup();let mut ev = vec![];for j in 0..std::cmp::min(i, n / 2) + 1 {for &(x, y) in &fst[j] {ev.push((l - x, 1, coo.lower_bound(&(p - y))));}}for &(x, y) in &snd[k - i] {ev.push((x, 0, coo.lower_bound(&y)));}ev.sort();let m = coo.len();let mut st = SegTree::new(m, |x, y| x + y, 0i64);for &(_, ty, idx) in &ev {if ty == 1 {ans += st.query(idx..m);} else {let old = st.query(idx..idx + 1);st.update(idx, old + 1);}}}println!("{}", ans);}