#851911 (Rust) No.2139 K Consecutive Sushi

• 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 (1.83.0 + proconio)
• 結果: AC
• 実行時間: 49 ms / 2,000 ms
• コード長: 3,297 bytes
• コンパイル時間: 11,791 ms
• コンパイル使用メモリ: 378,908 KB
• 実行使用メモリ: 10,624 KB
• 最終ジャッジ日時: 2024-11-27 19:55:48
• 合計ジャッジ時間: 13,913 ms
• ジャッジサーバーID （参考情報）: judge5 / judge1

ファイルパターン	結果
sample	AC * 3
other	AC * 31

ソースコード

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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]);
}
 
 
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