結果
問題 | No.2139 K Consecutive Sushi |
ユーザー | koba-e964 |
提出日時 | 2023-03-23 23:03:59 |
言語 | Rust (1.77.0 + proconio) |
結果 |
AC
|
実行時間 | 46 ms / 2,000 ms |
コード長 | 3,297 bytes |
コンパイル時間 | 11,867 ms |
コンパイル使用メモリ | 378,396 KB |
実行使用メモリ | 10,752 KB |
最終ジャッジ日時 | 2024-05-05 19:24:33 |
合計ジャッジ時間 | 13,742 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
5,248 KB |
testcase_01 | AC | 1 ms
5,248 KB |
testcase_02 | AC | 1 ms
5,376 KB |
testcase_03 | AC | 46 ms
10,624 KB |
testcase_04 | AC | 44 ms
10,624 KB |
testcase_05 | AC | 45 ms
10,624 KB |
testcase_06 | AC | 45 ms
10,624 KB |
testcase_07 | AC | 44 ms
10,496 KB |
testcase_08 | AC | 37 ms
10,624 KB |
testcase_09 | AC | 41 ms
10,624 KB |
testcase_10 | AC | 43 ms
10,624 KB |
testcase_11 | AC | 41 ms
10,624 KB |
testcase_12 | AC | 33 ms
10,624 KB |
testcase_13 | AC | 1 ms
5,376 KB |
testcase_14 | AC | 1 ms
5,376 KB |
testcase_15 | AC | 1 ms
5,376 KB |
testcase_16 | AC | 1 ms
5,376 KB |
testcase_17 | AC | 1 ms
5,376 KB |
testcase_18 | AC | 2 ms
5,376 KB |
testcase_19 | AC | 1 ms
5,376 KB |
testcase_20 | AC | 0 ms
5,376 KB |
testcase_21 | AC | 1 ms
5,376 KB |
testcase_22 | AC | 1 ms
5,376 KB |
testcase_23 | AC | 4 ms
5,376 KB |
testcase_24 | AC | 15 ms
5,376 KB |
testcase_25 | AC | 31 ms
9,088 KB |
testcase_26 | AC | 5 ms
5,376 KB |
testcase_27 | AC | 13 ms
5,376 KB |
testcase_28 | AC | 13 ms
5,376 KB |
testcase_29 | AC | 6 ms
5,376 KB |
testcase_30 | AC | 16 ms
5,632 KB |
testcase_31 | AC | 42 ms
10,112 KB |
testcase_32 | AC | 7 ms
5,376 KB |
testcase_33 | AC | 37 ms
10,752 KB |
ソースコード
use std::cmp::*; // https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 macro_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 ; $len:expr ]) => { (0..$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 2 SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e} } // ary[k] <- v pub 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) } } fn main() { input! { n: usize, k: usize, a: [i64; n], } const INF: i64 = 1 << 50; let mut st = SegTree::new(n + 1, max, -INF); let mut dp = vec![0; n + 2]; let mut acc = vec![0; n + 1]; for i in 0..n { acc[i + 1] = acc[i] + a[i]; } st.update(0, 0); st.update(1, -acc[1]); for i in 1..n + 1 { let me = st.query(max(i, k - 1) - (k - 1)..i) + acc[i]; dp[i + 1] = max(me, dp[i]); if i < n { st.update(i + 1, dp[i + 1] - acc[i + 1]); } } println!("{}", dp[n + 1]); }