AGC020 C - Median Sum

解法

解説の通り。

空でない部分列の数は  2^N -1 個存在するが、部分列の和が x となるような部分列の構成方法の数え上げは動的計画法 O(N^2 A) で求まる。だが、これでは間に合わない。

中央値は配列の合計の 1/2 以上になるということに気づくと、部分列の和が x となるような部分列が構成できるかどうかを判定すれば良いことになる。この DP は BitSet によって高速化出来るため、  O(N^2 A / 64) で間に合う。すごい。

コード

/// Thank you tanakh!!!
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!(n: usize, a: [usize; n]);
    let sum: usize = a.iter().sum();
    let mut dp = bitset::BitSet::new(sum + 1);
    dp.set(0, true);
    for i in 0..n {
        let pd = dp.shift_left(a[i]);
        dp |= pd;
    }

    for i in ((sum + 1) / 2)..(sum + 1) {
        if dp.get(i) {
            println!("{}", i);
            return;
        }
    }
}

pub mod bitset {
    use std::ops::{BitOrAssign, Shl};
    const ONE_VALUE_LENGTH: usize = 60;
    const MAXIMUM: u64 = (1 << ONE_VALUE_LENGTH) - 1;

    pub fn get_bit_position(index: usize) -> (usize, usize) {
        let data_index = index / ONE_VALUE_LENGTH;
        let bit_index = index % ONE_VALUE_LENGTH;
        (data_index, bit_index)
    }

    #[derive(PartialEq, Clone, Debug)]
    pub struct BitSet {
        data: Vec<u64>,
    }

    impl BitOrAssign for BitSet {
        fn bitor_assign(&mut self, rhs: Self) {
            if self.data.len() < rhs.data.len() {
                self.data.resize(rhs.data.len(), 0);
            }
            let n = if self.data.len() > rhs.data.len() {
                rhs.data.len()
            } else {
                self.data.len()
            };
            for i in 0..n {
                assert!(self.data[i] <= MAXIMUM);
                assert!(rhs.data[i] <= MAXIMUM);
                self.data[i] |= rhs.data[i];
            }
        }
    }

    impl Shl<usize> for BitSet {
        type Output = Self;
        fn shl(self, rhs: usize) -> Self {
            self.shift_left(rhs)
        }
    }

    impl BitSet {
        pub fn new(n: usize) -> Self {
            let size = (n + ONE_VALUE_LENGTH - 1) / ONE_VALUE_LENGTH;
            BitSet {
                data: vec![0; size],
            }
        }

        pub fn new_from(value: u64) -> Self {
            BitSet { data: vec![value] }
        }

        pub fn set(&mut self, index: usize, value: bool) {
            let (data_index, bit_index) = get_bit_position(index);
            assert!(self.data.len() > data_index);
            if value {
                self.data[data_index] |= 1 << bit_index;
            } else {
                let tmp = MAXIMUM ^ (1 << bit_index);
                self.data[data_index] &= tmp;
            }
        }

        pub fn get(&mut self, index: usize) -> bool {
            let (data_index, bit_index) = get_bit_position(index);
            assert!(self.data.len() > data_index);
            self.data[data_index] & (1 << bit_index) != 0
        }

        pub fn shift_left(&self, shift: usize) -> Self {
            let mut next_data = Vec::new();
            let prefix_empty_count = shift / ONE_VALUE_LENGTH;
            let shift_count = shift % ONE_VALUE_LENGTH;
            for _ in 0..prefix_empty_count {
                next_data.push(0);
            }

            let mut from_previous = 0;
            let room = ONE_VALUE_LENGTH - shift_count;
            for &data in self.data.iter() {
                let overflow = (data >> room) << room;
                let rest = data - overflow;
                let value = (rest << shift_count) + from_previous;
                assert!(value <= MAXIMUM);
                next_data.push(value);
                from_previous = overflow >> room;
            }
            if from_previous > 0 {
                next_data.push(from_previous);
            }
            BitSet { data: next_data }
        }
    }
}

みんなのプロコン2018 決勝 A- Uncommon

解法

ある整数 i について配列内の i と互いに素な整数の数を数えるのは大変そうなので、i と互いに素でない数を数えることにする。

i を素因数分解し、例えば素因数が p, q, r だったとき、配列 a に含まれる i と互いに素でないでない整数の数は以下のように求められる。

(配列 a に含まれる i と互いに素でないでない整数の数)
= (配列 a に含まれる p の倍数の数) + (配列 a に含まれる q の倍数の数) + (配列 a に含まれる r の倍数の数)
  - (配列 a に含まれる p*q の倍数の数) - (配列 a に含まれる q*r の倍数の数) - (配列 a に含まれる r*p の倍数の数)
  + (配列 a に含まれる p*q*r の倍数の数)

配列 a 内の中にある x の倍数の数をカウントしたいが、 \sum_{k=1}^N 1/k O(\log N) なので、以下のような愚直なループで求まる。

for x in 2..n:
  cur = x
  while cur <= max:
    sum[x] += count[cur]
    cur += x

コード

/// Thank you tanakh!!!
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn get_primes(n: usize) -> Vec<usize> {
    let mut is_prime = vec![true; n + 1];
    is_prime[0] = false;
    is_prime[1] = false;
    let mut primes = vec![];
    for i in 2..(n + 1) {
        if is_prime[i] {
            primes.push(i);
            let mut pos = i * 2;
            while pos < is_prime.len() {
                is_prime[pos] = false;
                pos += i;
            }
        }
    }
    primes
}

use std::collections::BTreeSet;

fn main() {
    input!(n: usize, m: usize, a: [usize; n]);

    let primes = get_primes(350);
    let mut divisors = vec![vec![]; m + 1];
    let mut set = BTreeSet::new();
    for i in 1..(m + 1) {
        let mut a = i;
        for &prime in primes.iter() {
            if a % prime == 0 {
                divisors[i].push(prime);
                set.insert(prime);
                while a % prime == 0 {
                    a /= prime;
                }
            }
            if prime * prime > a {
                break;
            }
        }
        if a > 1 {
            divisors[i].push(a);
            set.insert(a);
        }
    }

    let max: usize = *a.iter().max().unwrap();
    let mut a_count = vec![0; max + 1];
    for &a in a.iter() {
        a_count[a] += 1;
    }

    let mut divide_count = vec![0; m + 1];
    for i in 2..(m + 1) {
        let mut cur = i;
        let mut count = 0;
        while cur <= max {
            count += a_count[cur];
            cur += i;
        }
        divide_count[i] = count;
    }

    for i in 1..(m + 1) {
        let divisors = &divisors[i];
        let n = divisors.len();
        let mut ans: i64 = 0;
        for mask in 1..(1 << n) {
            let mut count_ones = 0;
            let mut t = 1;
            for i in 0..n {
                if mask & (1 << i) != 0 {
                    t *= divisors[i];
                    count_ones += 1;
                }
            }
            let sum = divide_count[t] as i64;
            let sum = if count_ones % 2 == 1 { sum } else { -sum };
            ans += sum;
        }
        println!("{}", a.len() as i64 - ans);
    }
}

AtCoder Regular Contest 101 D - Median of Medians

解法

完全に解法の通り。

  1. 全ての連続部分列のうち、中央値が x 以上であるものがいくつあるかを考える問題になる。
  2. 数列の中央値が x 以上である。 => 数列の各要素を x 以上なら 1 そうでなければ -1 に置き換えた時に数列の和が 0 以上である。
  3. 我が 0 以上になる連続部分列の数を求めたい。 => 累積和をとって転倒数を調べれば良い(これすごく頭いい変形だと思うけど典型っぽい)

コード

/// Thank you tanakh!!!
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

use std::collections::{BTreeMap, BTreeSet};
use std::ops::{AddAssign, Sub};

fn main() {
    input!(n: usize, a: [i64; n]);

    let mut ok = 0;
    let mut ng = 1e10 as i64;
    while ng - ok > 1 {
        let x = (ng + ok) / 2;
        let a = a
            .iter()
            .map(|&a| if a < x { -1 } else { 1 })
            .collect::<Vec<_>>();
        let mut sum = vec![0; n + 1];
        for i in 0..n {
            sum[i + 1] = sum[i] + a[i];
        }
        let mut set = BTreeSet::new();
        for i in 0..(n + 1) {
            set.insert(sum[i]);
        }
        let mut map = BTreeMap::new();
        for (i, &x) in set.iter().enumerate() {
            map.insert(x, i);
        }
        let mut compact = vec![0; n + 1];
        for i in 0..(n + 1) {
            compact[i] = *map.get(&sum[i]).unwrap();
        }

        let m = map.len();
        let mut ans = 0;
        let mut bit = FenwickTree::new(m, 0);
        for i in 0..(n + 1) {
            ans += bit.sum_one(compact[i] + 1);
            bit.add(compact[i], 1);
        }

        let total = n * (n - 1) / 2 + n;
        let half = (total + 1) / 2;
        if ans >= half {
            ok = x;
        } else {
            ng = x;
        }
    }
    println!("{}", ok);
}

pub struct FenwickTree<T> {
    n: usize,
    data: Vec<T>,
    init: T,
}

impl<T: Copy + AddAssign + Sub<Output = T>> FenwickTree<T> {
    /// Constructs a new `FenwickTree`. The size of `FenwickTree` should be specified by `size`.
    pub fn new(size: usize, init: T) -> FenwickTree<T> {
        FenwickTree {
            n: size + 1,
            data: vec![init; size + 1],
            init: init,
        }
    }

    pub fn add(&mut self, k: usize, value: T) {
        let mut x = k;
        while x < self.n {
            self.data[x] += value;
            x |= x + 1;
        }
    }

    /// Returns a sum of range `[l, r)`
    pub fn sum(&self, l: usize, r: usize) -> T {
        self.sum_one(r) - self.sum_one(l)
    }

    /// Returns a sum of range `[0, k)`
    pub fn sum_one(&self, k: usize) -> T {
        if k >= self.n {
            panic!("");
        }

        let mut result = self.init;
        let mut x = k as i32 - 1;
        while x >= 0 {
            result += self.data[x as usize];
            x = (x & (x + 1)) - 1;
        }

        result
    }
}

lowlink, 橋や関節点について練習

lowlink や橋や関節点について、いくつか問題を解いたのでリンクを貼っておく。

lowlink の説明

kagamiz.hatenablog.com

kagamiz 先生による有名記事。大変わかりやすい。

AOJ 3022: Cluster Network

解法

雰囲気から、関節点を求める機運が高まるが、実際の問題はその後に最後の答えを出力するパートである。

http://web-ext.u-aizu.ac.jp/circles/acpc/commentary/ACPC2017Day2/J.pdf

解説 PDF にあるとおり、 DFS Tree において、頂点 i と、その子 v について lowlink[v] < order[i] が成り立つならば、 v を根とする部分木と i の親は後退辺によって連結である、という性質を利用する。

コード

/// Thank you tanakh!!!
/// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

use std::cmp;
use std::collections::{BTreeSet, VecDeque};

const INF: usize = 1e15 as usize;

fn main() {
    input!(
        n: usize,
        m: usize,
        power: [usize; n],
        edges: [(usize1, usize1); m]
    );

    let mut graph = vec![vec![]; n];
    for &(u, v) in edges.iter() {
        graph[v].push(u);
        graph[u].push(v);
    }

    let bridge_detector = BridgeDetector::new(&graph);
    let total: usize = power.iter().sum();

    let mut is_articulation = vec![false; n];
    for &v in bridge_detector.articulations.iter() {
        is_articulation[v] = true;
    }

    let mut dp = vec![INF; n];
    let tree = bridge_detector.dfs_tree;
    let low_link = bridge_detector.low_link;
    let order = bridge_detector.order;
    for i in 0..n {
        if is_articulation[i] {
            let mut max_child = 0;
            let mut connected_to_parent = 0;
            for &child in tree[i].iter() {
                let dp_child = dfs(child, &tree, &mut dp, &power);
                max_child = cmp::max(max_child, dp_child);

                if low_link[child] < order[i] {
                    connected_to_parent += dp_child;
                }
            }

            let dp_parent = total - (dfs(i, &tree, &mut dp, &power) - connected_to_parent);

            println!("{}", cmp::max(dp_parent, max_child));
        } else {
            println!("{}", total - power[i]);
        }
    }
}

fn dfs(v: usize, tree: &Vec<Vec<usize>>, dp: &mut Vec<usize>, power: &Vec<usize>) -> usize {
    if dp[v] != INF {
        return dp[v];
    }

    dp[v] = power[v];
    for &next in tree[v].iter() {
        dp[v] += dfs(next, tree, dp, power);
    }
    dp[v]
}

pub struct BridgeDetector {
    articulations: Vec<usize>,
    bridges: Vec<(usize, usize)>,
    visit: Vec<bool>,
    order: Vec<usize>,
    low_link: Vec<usize>,
    k: usize,
    dfs_tree: Vec<Vec<usize>>,
}

impl BridgeDetector {
    pub fn new(graph: &Vec<Vec<usize>>) -> Self {
        let n = graph.len();
        let mut d = BridgeDetector {
            articulations: vec![],
            bridges: vec![],
            visit: vec![false; n],
            order: vec![0; n],
            low_link: vec![0; n],
            k: 0,
            dfs_tree: vec![vec![]; n],
        };
        d.run(graph);
        d
    }

    fn run(&mut self, graph: &Vec<Vec<usize>>) {
        let n = graph.len();
        for i in 0..n {
            if !self.visit[i] {
                self.dfs(i, 0, graph, i);
            }
        }
    }

    fn dfs(&mut self, v: usize, previous: usize, graph: &Vec<Vec<usize>>, root: usize) {
        self.visit[v] = true;
        self.order[v] = self.k;
        self.k += 1;
        self.low_link[v] = self.order[v];

        let mut is_articulation = false;
        let mut dimension = 0;
        for &next in graph[v].iter() {
            if !self.visit[next] {
                // The edge (v->next) is not a backedge.
                self.dfs_tree[v].push(next);
                dimension += 1;
                self.dfs(next, v, graph, root);
                self.low_link[v] = cmp::min(self.low_link[v], self.low_link[next]);
                if v != root && self.order[v] <= self.low_link[next] {
                    is_articulation = true;
                }
                if self.order[v] < self.low_link[next] {
                    let min = cmp::min(v, next);
                    let max = cmp::max(v, next);
                    self.bridges.push((min, max));
                }
            } else if v == root || next != previous {
                // The edge (v->next) is a backedge.
                self.low_link[v] = cmp::min(self.low_link[v], self.order[next]);
            }
        }

        if v == root && dimension > 1 {
            is_articulation = true;
        }
        if is_articulation {
            self.articulations.push(v);
        }
    }
}

ARC 039 D - 旅行会社高橋君

解法

橋を2回通らずに A->B->C の移動が出来るかという問題になる。

まず橋を検出し、それらを取り除くと、グラフはいくつかの連結成分に分かれる。各連結成分内では頂点間を移動する経路が常に複数あるため、移動に制限がかからない。これらの連結成分をそれぞれ頂点とし、元のグラフの橋を辺とした木を考える。この木の上で dist(A,B)+dist(B,C)==dist(C,A) ならば A->B->C は同じ橋を 2 回通らずに移動可能である。

コード

/// Thank you tanakh!!!
///  https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

use std::cmp;
use std::collections::{BTreeMap, BTreeSet, VecDeque};

fn main() {
    input!(
        n: usize,
        m: usize,
        edges: [(usize1, usize1); m],
        q: usize,
        queries: [(usize1, usize1, usize1); q]
    );
    solve(n, &edges, &queries);
}

fn solve(n: usize, edges: &Vec<(usize, usize)>, queries: &Vec<(usize, usize, usize)>) {
    let bridges = {
        let mut graph = vec![vec![]; n];
        for &(u, v) in edges.iter() {
            graph[u].push(v);
            graph[v].push(u);
        }

        let mut bridge_detector = BridgeDetector::new(n);
        bridge_detector.run(&graph);

        let mut bridges = BTreeSet::new();
        for &(a, b) in bridge_detector.bridges.iter() {
            bridges.insert((a, b));
            bridges.insert((b, a));
        }
        bridges
    };

    let mut uf = UnionFind::new(n);
    for &(a, b) in edges.iter() {
        if bridges.contains(&(a, b)) {
            continue;
        }
        uf.unite(a, b);
    }

    let mut set = BTreeSet::new();
    let mut find = vec![0; n];
    for i in 0..n {
        let f = uf.find(i);
        let cur = set.len();
        set.insert(f);
        if cur == set.len() {
            find[i] = find[f];
        } else {
            find[f] = cur;
            find[i] = find[f];
        }
    }

    let super_tree = {
        let mut super_tree = vec![vec![]; set.len()];
        for &(a, b) in edges.iter() {
            let a = find[a];
            let b = find[b];
            if a == b {
                continue;
            }
            super_tree[a].push(b);
            super_tree[b].push(a);
        }
        super_tree
    };

    let lca = LowestCommonAncestor::new(&super_tree);
    for &(a, b, c) in queries.iter() {
        let (a, b, c) = (find[a], find[b], find[c]);

        if lca.get_dist(a, b) + lca.get_dist(b, c) == lca.get_dist(a, c) {
            println!("OK");
        } else {
            println!("NG");
        }
    }
}

struct BridgeDetector {
    articulations: Vec<usize>,
    bridges: Vec<(usize, usize)>,
    visit: Vec<bool>,
    ord: Vec<usize>,
    low: Vec<usize>,
    k: usize,
}

impl BridgeDetector {
    fn new(n: usize) -> Self {
        BridgeDetector {
            articulations: vec![],
            bridges: vec![],
            visit: vec![false; n],
            ord: vec![0; n],
            low: vec![0; n],
            k: 0,
        }
    }

    fn run(&mut self, graph: &Vec<Vec<usize>>) {
        let n = graph.len();
        for i in 0..n {
            if !self.visit[i] {
                self.dfs(i, None, graph);
            }
        }
    }
    fn dfs(&mut self, v: usize, p: Option<usize>, graph: &Vec<Vec<usize>>) {
        self.visit[v] = true;
        self.ord[v] = self.k;
        self.k += 1;
        self.low[v] = self.ord[v];

        let mut is_articulation = false;
        let mut count = 0;
        for &next in graph[v].iter() {
            if !self.visit[next] {
                count += 1;
                self.dfs(next, Some(v), graph);
                if self.low[v] > self.low[next] {
                    self.low[v] = self.low[next];
                }
                if p.is_some() && self.ord[v] <= self.low[next] {
                    is_articulation = true;
                }
                if self.ord[v] < self.low[next] {
                    let (v, next) = if v < next { (v, next) } else { (next, v) };
                    self.bridges.push((v, next));
                }
            } else if p.is_none() || next != p.unwrap() && self.low[v] > self.ord[next] {
                self.low[v] = self.ord[next];
            }
        }

        if p.is_none() && count > 1 {
            is_articulation = true;
        }
        if is_articulation {
            self.articulations.push(v);
        }
    }
}

const MAX_PARENT: usize = 1 << 50;
pub struct LowestCommonAncestor {
    parent: Vec<Vec<usize>>,
    depth: Vec<usize>,
    log_v: usize,
}

impl LowestCommonAncestor {
    pub fn new(graph: &Vec<Vec<usize>>) -> Self {
        let num_v = graph.len();
        let root = 0;
        let mut depth = vec![0; num_v];

        let mut log_v = 1;
        let mut i = 1;
        while i <= num_v {
            i *= 2;
            log_v += 1;
        }
        let mut parent: Vec<Vec<usize>> = vec![vec![0; num_v]; log_v];

        let mut depth_vis = vec![false; num_v];
        let mut stack = VecDeque::new();
        stack.push_front(root);
        parent[0][root] = MAX_PARENT;
        depth[root] = 0;
        depth_vis[root] = true;
        while !stack.is_empty() {
            let v = stack.pop_front().unwrap();
            stack.push_front(v);
            for &u in &graph[v] {
                if depth_vis[u] {
                    continue;
                }
                parent[0][u] = v;
                depth[u] = depth[v] + 1;
                depth_vis[u] = true;
                stack.push_front(u);
            }

            let head = stack.pop_front().unwrap();
            if head != v {
                stack.push_front(head);
            }
        }

        for k in 0..(log_v - 1) {
            for u in 0..num_v {
                parent[k + 1][u] = if parent[k][u] == MAX_PARENT {
                    MAX_PARENT
                } else {
                    parent[k][parent[k][u]]
                };
            }
        }

        LowestCommonAncestor {
            parent: parent,
            depth: depth,
            log_v: log_v,
        }
    }

    pub fn get_lca(&self, u: usize, v: usize) -> usize {
        let (mut u, mut v) = if self.depth[u] <= self.depth[v] {
            (u, v)
        } else {
            (v, u)
        };
        for k in 0..self.log_v {
            if ((self.depth[v] - self.depth[u]) & (1 << k)) != 0 {
                v = self.parent[k][v];
            }
        }
        if u == v {
            return u;
        }

        for k in (0..self.log_v).rev() {
            if self.parent[k][u] != self.parent[k][v] {
                u = self.parent[k][u];
                v = self.parent[k][v];
            }
        }
        return self.parent[0][u];
    }

    pub fn get_dist(&self, u: usize, v: usize) -> usize {
        let lca = self.get_lca(u, v);
        self.depth[u] + self.depth[v] - self.depth[lca] * 2
    }
}

pub struct UnionFind {
    parent: Vec<usize>,
    sizes: Vec<usize>,
    size: usize,
}

impl UnionFind {
    pub fn new(n: usize) -> UnionFind {
        UnionFind {
            parent: (0..n).map(|i| i).collect::<Vec<usize>>(),
            sizes: vec![1; n],
            size: n,
        }
    }

    pub fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let px = self.parent[x];
            self.parent[x] = self.find(px);
            self.parent[x]
        }
    }

    pub fn unite(&mut self, x: usize, y: usize) -> bool {
        let parent_x = self.find(x);
        let parent_y = self.find(y);
        if parent_x == parent_y {
            return false;
        }

        let (large, small) = if self.sizes[parent_x] < self.sizes[parent_y] {
            (parent_y, parent_x)
        } else {
            (parent_x, parent_y)
        };

        self.parent[small] = large;
        self.sizes[large] += self.sizes[small];
        self.sizes[small] = 0;
        self.size -= 1;
        return true;
    }
}

CS Academy: Growing Segment

問題と解法

記憶力があれなので同じ問題を何度も解く。

kenkoooo.hatenablog.com

コード

use std::cmp;
use std::collections::BTreeSet;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let mut s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input!{
        n: usize,
        q: usize,
        x: [i64; n],
        l: [i64; q],
    }
    let mut l: Vec<(i64, usize)> = (0..q).map(|i| (l[i], i)).collect();
    l.sort();

    let mut x = {
        let mut v: Vec<i64> = vec![0];
        for &x in x.iter() {
            while v.len() >= 2 && (v[v.len() - 2] < v[v.len() - 1]) == (v[v.len() - 1] < x) {
                v.pop();
            }
            v.push(x);
        }
        v
    };

    let mut offset = 0;
    if x.len() >= 2 && x[1] < 0 {
        offset = -x[1];
        x.remove(0);
        for x in x.iter_mut() {
            *x += offset;
        }
    }

    let mut point_set = BTreeSet::new();
    for i in 0..x.len() {
        point_set.insert(i);
    }

    let mut edge_set = EdgeSet::new(offset);
    for i in 1..x.len() {
        edge_set.insert(i - 1, i, &x);
    }

    let mut ans = vec![0; q];
    for &(length, ans_id) in l.iter() {
        while let Some(&(distance, i1, i2)) = edge_set.first() {
            if distance >= length {
                break;
            }

            if edge_set.set.len() == 1 {
                break;
            }

            if i1 == 0 {
                let &i3 = point_set.range((i2 + 1)..).next().unwrap();

                edge_set.remove(i1, i2, &x);
                edge_set.remove(i2, i3, &x);

                x[i2] = length;
                edge_set.insert(i1, i2, &x);
                edge_set.insert(i2, i3, &x);
                continue;
            }

            let i3 = point_set.range((i2 + 1)..).next().cloned();
            if i3.is_none() {
                edge_set.remove(i1, i2, &x);
                point_set.remove(&i2);
                continue;
            }

            let &i0 = point_set.range(..i1).next_back().unwrap();
            let i3 = i3.unwrap();
            assert!((x[i0] < x[i1]) != (x[i1] < x[i2]));
            assert!((x[i1] < x[i2]) != (x[i2] < x[i3]));

            // remove i2
            edge_set.remove(i1, i2, &x);
            edge_set.remove(i2, i3, &x);
            point_set.remove(&i2);

            if (x[i0] - x[i1]).abs() > (x[i0] - x[i3]).abs() {
                // remove i3
                let &i4 = point_set.range((i3 + 1)..).next_back().unwrap();
                point_set.remove(&i3);
                edge_set.remove(i3, i4, &x);
                edge_set.insert(i1, i4, &x);
            } else {
                // remove i1
                point_set.remove(&i1);
                edge_set.remove(i0, i1, &x);
                edge_set.insert(i0, i3, &x);
            }
        }

        ans[ans_id] = cmp::max(0, edge_set.total - (edge_set.set.len() as i64) * length);
    }
    for &ans in ans.iter() {
        println!("{}", ans);
    }
}

#[derive(Debug)]
struct EdgeSet {
    set: BTreeSet<(i64, usize, usize)>,
    total: i64,
}

impl EdgeSet {
    fn new(offset: i64) -> Self {
        EdgeSet {
            set: BTreeSet::new(),
            total: offset,
        }
    }

    fn insert(&mut self, from: usize, to: usize, x: &Vec<i64>) {
        assert!(from < to);
        let distance = (x[from] - x[to]).abs();
        self.set.insert((distance, from, to));
        self.total += distance;
    }

    fn remove(&mut self, from: usize, to: usize, x: &Vec<i64>) {
        assert!(from < to);
        let distance = (x[from] - x[to]).abs();
        self.set.remove(&(distance, from, to));
        self.total -= distance;
    }

    fn first(&self) -> Option<&(i64, usize, usize)> {
        self.set.iter().next()
    }
}