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[番外編] アルゴリズム・データ構造ごとに問題を分類してみる その2

競プロで学ぶRust

アルゴリズムやデータ構造ごとに解ける問題を分類しました。

アルゴリズムやデータ構造ごとに問題を分類する その2

目次

アルゴリズム
幅優先探索-7問
ダイクストラ法-6問
半分全列挙

アルゴリズム

幅優先探索-7問

ABC400 D - Takahashi the Wall Breaker

D - Takahashi the Wall Breaker（Difficulty : 1026）

典型的な01-BFS。

コード例を見る 
use std::collections::VecDeque;

fn out_of_bounds(h: usize, w: usize, i: isize, j: isize) -> bool {
    i < 0 || j < 0 || i == h as isize || j == w as isize
}

fn run(h: usize, w: usize, s: Vec<&str>, a: usize, b: usize, c: usize, d: usize) -> usize {
    let vec: Vec<Vec<char>> = s.into_iter().map(|s| s.chars().collect()).collect();

    let mut dist = vec![vec![-1; w]; h];
    dist[a - 1][b - 1] = 0;

    let mut queue = VecDeque::new();
    queue.push_back((a - 1, b - 1));

    let di = [0, 1, 0, -1];
    let dj = [1, 0, -1, 0];

    while let Some((cur_i, cur_j)) = queue.pop_front() {
        for i in 0..4 {
            let new_i = cur_i as isize + di[i];
            let new_j = cur_j as isize + dj[i];

            if out_of_bounds(h, w, new_i, new_j) {
                continue;
            }

            let new_i = new_i as usize;
            let new_j = new_j as usize;

            if vec[new_i][new_j] != '#' {
                if dist[new_i][new_j] == -1 || dist[new_i][new_j] > dist[cur_i][cur_j] {
                    dist[new_i][new_j] = dist[cur_i][cur_j];
                    queue.push_front((new_i, new_j));
                }
            } else {
                // 1マス先
                if dist[new_i][new_j] == -1 {
                    dist[new_i][new_j] = dist[cur_i][cur_j] + 1;
                    queue.push_back((new_i, new_j));
                }

                // 2マス先の座標
                let new_i2 = cur_i as isize + dx[i] * 2;
                let new_j2 = cur_j as isize + dy[i] * 2;

                if out_of_bounds(h, w, new_i2, new_j2) {
                    continue;
                }

                let new_i2 = new_i2 as usize;
                let new_j2 = new_j2 as usize;

                if dist[new_i2][new_j2] != -1 {
                    continue;
                }

                dist[new_i2][new_j2] = dist[cur_i][cur_j] + 1;
                queue.push_back((new_i2, new_j2));
            }
        }
    }

    dist[c - 1][d - 1] as usize
}

ABC254 E - Small d and k

E - Small d and k（Difficulty : 1202）

コード例を見る 
// https://atcoder.jp/contests/abc254/tasks/abc254_e

use std::collections::{HashMap, VecDeque};

fn run(n: usize, _m: usize, ab: Vec<(usize, usize)>, _q: usize, xk: Vec<(usize, usize)>) -> Vec<usize> {
    let mut hash_map = HashMap::new();

    for (a, b) in ab {
        hash_map.entry(a).or_insert_with(Vec::new).push(b);
        hash_map.entry(b).or_insert_with(Vec::new).push(a);
    }

    let mut ans = Vec::new();

    for (x, k) in xk {
        if let None =  hash_map.get(&x) {
            ans.push(x);
            continue;
        }

        let mut graph = vec![false; n];
        let mut queue = VecDeque::new();

        queue.push_back((x, k));

        // 辿ったノードの合計
        let mut sum = x;

        while let Some((x, k)) = queue.pop_front() {
            if k == 0 {
                continue;
            }

            graph[x-1] = true;

            let next = hash_map.get(&x).unwrap();

            for n in next {
                if !graph[n-1] {
                    graph[n-1] = true;
                    queue.push_back((*n, k-1));
                    sum += *n;
                }
            }
        }

        ans.push(sum);
    }

    ans
}

ABC213 E - Stronger Takahashi

E - Stronger Takahashi（Difficulty : 1423）

これも01-BFS。

コード例を見る 
use std::collections::VecDeque;

fn out_of_bounds(h: usize, w: usize, i: isize, j: isize) -> bool {
    i < 0 || j < 0 || h as isize <= i || w as isize <= j
}

fn run(h: usize, w: usize, s: Vec<&str>) -> usize {
    let vec: Vec<Vec<char>> = s.into_iter().map(|str| str.chars().collect()).collect();

    let mut dist = vec![vec![std::usize::MAX; w]; h];
    dist[0][0] = 0;

    let mut queue = VecDeque::new();
    queue.push_back((0, 0));

    let di = [0, 1, 0, -1];
    let dj = [1, 0, -1, 0];

    while let Some((cur_i, cur_j)) = queue.pop_front() {
        for i in 0..4 {
            let new_i = cur_i as isize + di[i];
            let new_j = cur_j as isize + dj[i];

            if out_of_bounds(h, w, new_i, new_j) {
                continue;
            }

            let new_i = new_i as usize;
            let new_j = new_j as usize;

            if vec[new_i][new_j] == '#' {
                continue;
            }

            if dist[new_i][new_j] > dist[cur_i][cur_j] {
                dist[new_i][new_j] = dist[cur_i][cur_j];
                queue.push_front((new_i, new_j));
            }
        }

        for di in -2..=2 {
            for dj in -2..=2 {
                let new_i = cur_i as isize + di;
                let new_j = cur_j as isize + dj;

                if di.abs() + dj.abs() == 4 {
                    continue;
                }

                if out_of_bounds(h, w, new_i, new_j) {
                    continue;
                }

                let new_i = new_i as usize;
                let new_j = new_j as usize;

                if dist[new_i][new_j] > dist[cur_i][cur_j] + 1 {
                    dist[new_i][new_j] = dist[cur_i][cur_j] + 1;
                    queue.push_back((new_i, new_j));
                }
            }
        }
    }

    dist[h-1][w-1]
}

ABC020 C - 壁抜け

C - 壁抜け（🧪 Difficulty : 1477）

BFSと解の二分探索の組み合わせ。

コード例を見る 
use std::collections::VecDeque;

fn out_of_bounds(h: usize, w: usize, i: isize, j: isize) -> bool {
    i < 0 || j < 0 || h as isize == i || w as isize == j
}

fn bfs(h: usize, w: usize, t: usize, s: Vec<&str>, x: isize) -> bool {
    let vec: Vec<Vec<char>> = s.iter().map(|str| str.chars().collect()).collect();

    let mut s = (0, 0);
    let mut g = (0, 0);

    for i in 0..h {
        for j in 0..w {
            if vec[i][j] == 'S' {
                s = (i, j);
            }

            if vec[i][j] == 'G' {
                g = (i, j);
            }
        }
    }

    let mut dist = vec![vec![std::isize::MAX; w]; h];
    dist[s.0][s.1] = 0;

    let mut queue = VecDeque::new();
    queue.push_back(s);

    let di = [0, 1, 0, -1];
    let dj = [1, 0, -1, 0];

    while let Some((cur_i, cur_j)) = queue.pop_front() {
        for i in 0..4 {
            let new_i = cur_i as isize + di[i];
            let new_j = cur_j as isize + dj[i];

            if out_of_bounds(h, w, new_i, new_j) {
                continue;
            }

            let new_i = new_i as usize;
            let new_j = new_j as usize;

            let cost = if vec[new_i][new_j] == '#' { x } else { 1 };

            if dist[new_i][new_j] > dist[cur_i][cur_j] + cost {
                dist[new_i][new_j] = dist[cur_i][cur_j] + cost;

                if cost == 1 {
                    queue.push_front((new_i, new_j));
                } else {
                    queue.push_back((new_i, new_j));
                }
            }
        }
    }

    dist[g.0][g.1] <= t as isize
}

fn run(h: usize, w: usize, t: usize, s: Vec<&str>) -> usize {
    let mut low = 1;
    let mut high = 1_000_000_000;
    let mut ans = 1;

    while low <= high {
        let mid = (low + high) / 2;

        if bfs(h, w, t, s.clone(), mid) {
            ans = mid;
            low = mid + 1;
        } else {
            high = mid - 1;
        }
    }

    ans as usize
}

ABC176 D - Wizard in Maze

D - Wizard in Maze（Difficulty : 1276）

01-BFSその3。

コード例を見る 
use std::collections::VecDeque;

fn check(i: isize, j: isize, h: isize, w: isize) -> bool {
    i < 0 || j < 0 || i >= h || j >= w
}

const INF: isize = std::isize::MAX;

fn run(h: usize, w: usize, c: (usize, usize), d: (usize, usize), s: Vec<&str>) -> isize {
    let vec: Vec<Vec<char>> = s.into_iter().map(|s| s.chars().collect()).collect();

    let mut dist = vec![vec![INF; w]; h];
    dist[c.0 - 1][c.1 - 1] = 0;

    let mut queue = VecDeque::new();
    queue.push_back((c.0-1, c.1-1));

    let dx = [0, 1, 0, -1];
    let dy = [1, 0, -1, 0];

    while let Some((cur_i, cur_j)) = queue.pop_front() {
        for i in 0..4 {
            if check(cur_i as isize + dx[i], cur_j as isize + dy[i], h as isize, w as isize) {
                continue;
            }

            let next_i = (cur_i as isize + dx[i]) as usize;
            let next_j = (cur_j as isize + dy[i]) as usize;

            if vec[next_i][next_j] == '#' || dist[next_i][next_j] <= dist[cur_i][cur_j] {
                continue;
            }

            dist[next_i][next_j] = dist[cur_i][cur_j];
            queue.push_front((next_i, next_j));
        }

        for i in -2..=2 {
            for j in -2..=2 {
                let new_i = cur_i as isize + i;
                let new_j = cur_j as isize + j;

                if check(cur_i as isize + i, cur_j as isize + j, h as isize, w as isize) {
                    continue;
                }

                let jump_i = new_i as usize;
                let jump_j = new_j as usize;

                if vec[jump_i][jump_j] == '#' || dist[jump_i][jump_j] <= dist[cur_i][cur_j] + 1 {
                    continue;
                }

                dist[jump_i][jump_j] = dist[cur_i][cur_j] + 1;
                queue.push_back((jump_i, jump_j));
            }
        }
    }

    if dist[d.0-1][d.1-1] == INF {
        -1
    } else {
        dist[d.0-1][d.1-1]
    }
}

ARC005 C - 器物損壊！高橋君

C - 器物損壊！高橋君（🧪 Difficulty : 1503）

01-BFSその4。

コード例を見る 
use std::collections::VecDeque;

fn check(i: isize, j: isize, h: isize, w: isize) -> bool {
    i < 0 || j < 0 || i >= h || j >= w
}

fn run(h: usize, w: usize, c: Vec<&str>) -> &'static str {
    let vec: Vec<Vec<char>> = c.into_iter().map(|s| s.chars().collect()).collect();

    let mut s = (0, 0);
    let mut g = (0, 0);

    for i in 0..h {
        for j in 0..w {
            if vec[i][j] == 's' {
                s = (i, j);
            }

            if vec[i][j] == 'g' {
                g = (i, j);
            }

        }
    }

    let mut dist = vec![vec![-1; w]; h];
    dist[s.0][s.1] = 0;

    let mut queue = VecDeque::new();
    queue.push_back((s.0, s.1));

    let dx = [0, 1, 0, -1];
    let dy = [1, 0, -1, 0];

    while let Some((cur_i, cur_j)) = queue.pop_front() {
        for i in 0..4 {
            if check(cur_i as isize + dx[i],cur_j as isize + dy[i], h as isize, w as isize) {
                continue;
            }

            let next_i = (cur_i as isize + dx[i]) as usize;
            let next_j = (cur_j as isize + dy[i]) as usize;

            if dist[next_i][next_j] != -1 {
                continue;
            }

            if vec[next_i][next_j] != '#' {
                dist[next_i][next_j] = dist[cur_i][cur_j];
                queue.push_front((next_i, next_j));
            } else {
                dist[next_i][next_j] = dist[cur_i][cur_j] + 1;
                queue.push_back((next_i, next_j));
            }
        }
    }

    if dist[g.0][g.1] <= 2 {
        "YES"
    } else {
        "NO"
    }
}

ABC218 F - Blocked Roads

F - Blocked Roads（Difficulty : 1753）

最初にBFSで最短経路を求めておく。後は各エッジについて最短経路に含まれないならdist[n]を出力、最短経路に含まれるなら再度BFSを回すだけで解ける。

コード例を見る 
use std::collections::{HashMap, VecDeque};

fn bfs(n: usize, start: usize, hash_map: &HashMap<usize, Vec<usize>>) -> Vec<usize> {
    let mut dist = vec![std::usize::MAX; n+1];
    dist[start] = 0;

    let mut queue = VecDeque::new();
    queue.push_front(start);

    while let Some(cur) = queue.pop_front() {
        let Some(next) = hash_map.get(&cur) else {
            continue;
        };

        for next in next {
            if dist[*next] != std::usize::MAX {
                continue;
            }

            dist[*next] = dist[cur] + 1;

            queue.push_back(*next);
        }
    }

    dist
}

fn run(n: usize, _m: usize, st: Vec<(usize, usize)>) -> Vec<isize> {
    let mut graph = HashMap::new();

    for &(s, t) in &st {
        graph.entry(s).or_insert_with(Vec::new).push(t);
    }

    let dist = bfs(n, 1, &graph);

    let mut ans = Vec::new();

    for &(s, t) in &st {
        if dist[s] + 1 != dist[t] {
            ans.push(dist[n] as isize);
            continue;
        }

        let mut graph_clone = graph.clone();
        graph_clone.get_mut(&s).map(|v| v.retain(|&x| x != t));

        let dist_after_removal = bfs(n, 1, &graph_clone);

        let dist: isize = if dist_after_removal[n] == std::usize::MAX {
            -1
        } else {
            dist_after_removal[n] as isize
        };

        ans.push(dist);
    }

    ans
}

ダイクストラ法-6問

競技プログラミングの鉄則 A64 - Shortest Path 2

A64 - Shortest Path 2（Difficultyなし）

コード例を見る 
use std::collections::{BinaryHeap, HashMap};
use std::cmp::Reverse;

const INF: usize = std::usize::MAX;

fn run(n: usize, _m: usize, abc: Vec<(usize, usize, usize)>) -> Vec<isize> {
    let mut hash_map = HashMap::new();

    for (a, b, c) in abc {
        hash_map.entry(a).or_insert_with(|| Vec::new()).push((c, b));
        hash_map.entry(b).or_insert_with(|| Vec::new()).push((c, a));
    }

    let mut current = vec![INF; n+1];

    current[1] = 0;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push(Reverse((0, 1)));


    while let Some(Reverse((cur_cost, cur_i))) = priority_queue.pop() {
        if cur_cost > current[cur_i] {
            continue;
        }

        let Some(next) = hash_map.get(&cur_i) else {
            continue;
        };

        for (next_cost, next_i) in next {
            let new_cost = cur_cost + next_cost;

            if new_cost < current[*next_i] {
                current[*next_i] = new_cost;
                priority_queue.push(Reverse((new_cost, *next_i)));
            }
        }
    }

    current[1..].into_iter()
        .map(|c| {
            if *c == INF {
                -1
            } else {
                *c as isize
            }
        })
        .collect()
}

ABC012 D バスと避けられない運命

D - バスと避けられない運命（🧪 Difficulty : 1166）

コード例を見る 
use std::cmp::{min, Reverse};
use std::collections::{BinaryHeap, HashMap};

fn dijkstra(n: usize, start: usize, hash_map: &HashMap<usize, Vec<(usize, usize)>>) -> usize {
    let mut dist = vec![std::usize::MAX; n+1];
    dist[start] = 0;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push(Reverse((0, start)));

    while let Some(Reverse((cur_cost, cur_i))) = priority_queue.pop() {
        for (next_cost, next_i) in hash_map.get(&cur_i).unwrap() {
            let new_cost = cur_cost + next_cost;

            if new_cost < dist[*next_i] {
                dist[*next_i] = new_cost;
                priority_queue.push(Reverse((new_cost, *next_i)));
            }
        }
    }

    dist.into_iter().filter(|n| *n != std::usize::MAX).max().unwrap()
}

fn run(n: usize, _m: usize, abt: Vec<(usize, usize, usize)>) -> usize {
    let mut ans = std::usize::MAX;

    let mut hash_map = HashMap::new();

    for (a, b, t) in abt {
        hash_map.entry(a).or_insert_with(|| Vec::new()).push((t, b));
        hash_map.entry(b).or_insert_with(|| Vec::new()).push((t, a));
    }

    for i in 1..=n {
        ans = min(ans, dijkstra(n, i, &hash_map));
    }

    ans
}

ABC325 E - Our clients, please wait a moment

E - Our clients, please wait a moment（Difficulty : 1093）

コード例を見る 
use std::collections::{BinaryHeap, HashMap};
use std::cmp::{min, Reverse};

const INF: usize = std::usize::MAX;

fn dijkstra(
    n: usize,
    start: usize,
    a: usize,
    b: usize,
    c: usize,
    d: &Vec<Vec<usize>>,
    is_forward: bool
) -> Vec<usize> {
    let mut dist = vec![INF; n+1];

    dist[start] = 0;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push(Reverse((0, start)));

    while let Some(Reverse((cur_cost, cur_i))) = priority_queue.pop() {
        if cur_cost > dist[cur_i] {
            continue;
        }

        for next in 1..=n {
            if next == cur_i {
                continue;
            }

            let new_cost =
                if is_forward {
                    cur_cost + d[cur_i-1][next-1] * a
                } else {
                    cur_cost + (d[cur_i-1][next-1] * b) + c
                };

            if new_cost < dist[next] {
                dist[next] = new_cost;
                priority_queue.push(Reverse((new_cost, next)));
            }
        }
    }

    dist
}

fn run(n: usize, a: usize, b: usize, c: usize, d: Vec<Vec<usize>>) -> usize {
    let dijk = dijkstra(n, 1, a, b, c, &d, true);
    let dijk2 = dijkstra(n, n, a, b, c, &d, false);

    (1..=n)
        .map(|i| dijk[i] + dijk2[i])
        .min()
        .unwrap()
}

ABC335 E - Non-Decreasing Colorful Path

E - Non-Decreasing Colorful Path（Difficulty : 1540）

コード例を見る 
use std::collections::{BinaryHeap, HashMap};
use std::cmp::Reverse;

fn dijkstra(
    n: usize,
    hash_map: &HashMap<usize, Vec<usize>>,
    a: &Vec<usize>
) -> usize {
    let mut count = vec![0; n+1];
    count[1] = 1;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push((Reverse(a[0]), 1, 1));

    while let Some((_, cur_count, cur_i)) = priority_queue.pop() {
        if count[cur_i] > cur_count {
            continue;
        }

        let Some(next) = hash_map.get(&cur_i) else {
            continue;
        };

        for next_i in next {
            if a[cur_i-1] > a[*next_i-1] {
                continue;
            }

            let new_count = if a[cur_i - 1] < a[next_i - 1] {
                cur_count + 1
            } else {
                cur_count
            };

            if count[*next_i] < new_count {
                count[*next_i] = new_count;
                priority_queue.push((Reverse(a[*next_i-1]), new_count, *next_i));
            }
        }
    }

    count[n]
}

fn run(
    n: usize,
    _m: usize,
    a: Vec<usize>,
    uv: Vec<(usize, usize)>
) -> usize {
    let mut hash_map = HashMap::new();

    for (u, v) in uv {
        hash_map.entry(u).or_insert_with(|| Vec::new()).push(v);
        hash_map.entry(v).or_insert_with(|| Vec::new()).push(u);
    }

    dijkstra(n, &hash_map, &a)
}

ABC051 D - Candidates of No Shortest Paths

D - Candidates of No Shortest Paths（Difficulty : 1566）

コード例を見る 
use std::collections::{BinaryHeap, HashMap};

const INF: usize = std::usize::MAX;

fn dijkstra(n: usize, start: usize, hash_map: &HashMap<usize, Vec<(usize, usize)>>) -> Vec<usize> {
    let mut dist = vec![INF; n+1];
    dist[start] = 0;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push((0, start));

    while let Some((cur_cost, cur_i)) = priority_queue.pop() {
        let Some(next) = hash_map.get(&cur_i) else {
            continue;
        };

        for (next_cost, next_i) in next {
            let new_cost = cur_cost + next_cost;

            if new_cost < dist[*next_i] {
                dist[*next_i] = new_cost;
                priority_queue.push((new_cost, *next_i));
            }
        }
    }

    dist
}

fn run(n: usize, m: usize, abc: Vec<(usize, usize, usize)>) -> usize {
    let mut hash_map = HashMap::new();

    for &(a, b, c) in &abc {
        hash_map.entry(a).or_insert_with(|| Vec::new()).push((c, b));
        hash_map.entry(b).or_insert_with(|| Vec::new()).push((c, a));
    }

    let mut dist = vec![vec![INF; n + 1]; n + 1];

    for i in 1..=n {
        dist[i] = dijkstra(n, i, &hash_map);
    }

    let mut used = vec![false; m];

    for i in 1..=n {
        for j in 1..=n {
            if i == j || dist[i][j] == INF {
                continue;
            }

            for (k, &(a, b, c)) in abc.iter().enumerate() {
                if dist[i][a] + c + dist[b][j] == dist[i][j] || dist[i][b] + c + dist[a][j] == dist[i][j] {
                    used[k] = true;
                }
            }
        }
    }

    used.iter().filter(|&&u| !u).count()
}

ABC035 D - トレジャーハント

D - トレジャーハント（🧪 Difficulty : 1591）

コード例を見る 
use std::collections::{BinaryHeap, HashMap};
use std::cmp::Reverse;

const INF: usize = std::usize::MAX;

fn dijkstra(n: usize, start: usize, hash_map: &HashMap<usize, Vec<(usize, usize)>>) -> Vec<usize> {
    let mut dist = vec![INF; n+1];
    dist[start] = 0;

    let mut priority_queue = BinaryHeap::new();
    priority_queue.push(Reverse((0, start)));

    while let Some(Reverse((cur_cost, cur_i))) = priority_queue.pop() {
        if cur_cost > dist[cur_i] {
            continue;
        }

        if let Some(next) = hash_map.get(&cur_i) {
            for &(next_cost, next_i) in next {
                let new_cost = next_cost + cur_cost;

                if new_cost < dist[next_i] {
                    dist[next_i] = new_cost;
                    priority_queue.push(Reverse((new_cost, next_i)));
                }
            }
        }
    }

    dist
}

fn run(n: usize, _m: usize, t: usize, a: Vec<usize>, abc: Vec<(usize, usize, usize)>) -> usize {
    let mut forward = HashMap::new();
    let mut backward = HashMap::new();

    for (a, b, c) in abc {
        forward.entry(a).or_insert_with(Vec::new).push((c, b));
        backward.entry(b).or_insert_with(Vec::new).push((c, a));
    }

    let forward = dijkstra(n, 1, &forward);
    let backward = dijkstra(n, 1, &backward);

    let mut ans = 0;

    for i in 1..=n {
        if forward[i] == INF || backward[i] == INF {
            continue;
        }

        // 往復時間がT秒未満なら
        if forward[i] + backward[i] < t {
            ans = ans.max((t - forward[i] - backward[i]) * a[i-1]);
        }
    }

    ans
}

半分全列挙

ABC292 C - Four Variables

C - Four Variables（Difficulty : 444）

コード例を見る 
fn run(n: usize) -> usize {
    let mut ab = vec![0; n+1];

    for i in 1..=n {
        for j in 1..=(n / i) {
            ab[i*j] += 1;
        }
    }

    (1..=n)
        .map(|i| {
            ab[i] * ab[n-i]
        })
        .sum()
}

ABC143 D - Triangles

D - Triangles（Difficulty : 686）

コード例を見る 
use std::cmp::Ordering;
use itertools::Itertools;

fn upper_bound<T: Ord>(vec: &[T], value: T) -> usize {
    vec.binary_search_by(|x| {
        if *x <= value {
            Ordering::Less
        } else {
            Ordering::Greater
        }
    })
    .err()
    .unwrap()
}


fn run(n: usize, l: Vec<usize>) -> usize {
    let vec: Vec<usize> = l.into_iter().sorted().collect();

    let mut ans = 0;

    for i in 0..n {
        for j in i+1..n {
            let res = upper_bound(&vec, vec[i] + vec[j] - 1);

            if res > j + 1 {
                ans += res - j - 1;
            }
        }
    }

    ans
}

ABC184 F - Programming Contest

F - Programming Contest（Difficulty : 1432）

コード例を見る 
use itertools::Itertools;
use std::cmp::Ordering;

// upper_boundの拡張
// n以下の最大の数を返す
fn max_under_n<T: Ord>(vec: &[T], value: T) -> Option<usize> {
    vec.binary_search_by(|x| {
        if *x <= value {
            Ordering::Less
        } else {
            Ordering::Greater
        }
    })
    .err()
    .map(|x| if x == 0 {
        None
    } else {
        Some(x - 1)
    })
    .flatten()
}

fn run(n: usize, t: usize, a: Vec<usize>) -> usize {
    let (l, r) = a.split_at(n/2);

    let mut p = Vec::new();
    let mut q = Vec::new();

    for i in 0..=l.len() {
        for combination in l.iter().combinations(i) {
            let sum: usize = combination.iter().map(|&&x| x).sum();
            p.push(sum);
        }
    }

    for i in 0..=r.len() {
        for combination in r.iter().combinations(i) {
            let sum: usize = combination.iter().map(|&&x| x).sum();
            q.push(sum)
        }
    }

    q.sort();

    let mut ans = 0;

    for left in p.iter() {
        if t < *left {
            continue;
        }

        if let Some(right_idx) = max_under_n(&q, t - left) {
            ans = ans.max(q[right_idx] + left)
        }
    }

    ans
}

ABC123 D - Cake 123

D - Cake 123（Difficulty : 1489）

コード例を見る 
use std::collections::BinaryHeap;
use std::cmp::Reverse;

fn run(_x: usize, _y: usize, _z: usize, k: usize, a: Vec<usize>, b: Vec<usize>, c: Vec<usize>) -> Vec<usize> {
    let mut ab = BinaryHeap::new();

    // AとBの和で大きい順にK個求める
    for i in a.iter() {
        for j in b.iter() {
            ab.push(Reverse(i + j));

            if ab.len() > k {
                ab.pop();
            }
        }
    }

    let mut vec: Vec<_> = ab.into_sorted_vec();
    vec.reverse();

    let mut ans = BinaryHeap::new();

    // ABとCの和で大きい順にK個求める
    for i in vec.iter() {
        for j in c.iter() {
            ans.push(Reverse(i.0 + j));

            if ans.len() > k {
                ans.pop();
            }
        }
    }

    ans.into_sorted_vec()
        .into_iter()
        .map(|x| x.0)
        .collect()
}
更新履歴
  • 2025年06月08日 : ABC213 E - Stronger Takahashiを追加
  • 2025年06月07日 : ABC020 🧪 C - 壁抜けを追加
  • 2025年04月12日 : ABC400 D - Takahashi the Wall Breakerを追加
  • 2025年03月22日 : ABC218 F - Blocked Roadsを追加
  • 2025年03月09日 : ABC012 🧪 D - バスと避けられない運命を追加
  • 2025年03月03日 : ABC176 D - Wizard in Mazeを追加
  • 2025年03月02日 : ARC005 🧪 C - 器物損壊！高橋君を追加
  • 2025年03月01日 : ABC035 🧪 D - トレジャーハントを追加
  • 2025年02月28日 : ABC335 E - Non-Decreasing Colorful Pathを追加
  • 2025年02月23日 : ABC325 E - Our clients, please wait a momentを追加
  • 2025年01月13日 : ABC254 E - Small d and kを追加

参考