莫号模板库

杂项

约定

  1. C++ 版本：
    1. 最低支持版本：c++17；
    2. 默认版本：gnu++17；
  2. 格式：
    1. 缩进宽度为 4 个空格；
    2. 大括号换行；
    3. 一元运算符不空格，其他运算符空一格；
  3. 命名：
    1. 遵循 GoLang 的命名规范；
  4. 其他：
    1. 使用邻接表存图；
    2. 使用 0-indexed 的左闭右开区间；
    3. 使用解绑同步流的 std::cin 和 std::cout 输入输出；
    4. 使用 using namespace std; 和 #define int long long；
    5. 使用局部变量而非全局变量，局部 lambda 而非全局函数；
    6. 使用 emplace 而非 push，集合查重使用 count；
    7. 若键值较小，使用 vector 而非 map；不要求顺序的情况下可以用 unordered_ 系列，但在 Codeforces 上记得使用随机模数；

初始代码

#include <bits/stdc++.h>
using namespace std;

#define int long long

int T{0};
void solve() {}

signed main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    if (!T)
        cin >> T;
    while (T--)
        solve();
}

类型定义

using i8 = signed char;
using u8 = unsigned char;
using i32 = signed;
using u32 = unsigned;
using i64 = int64_t;
using u64 = uint64_t;
using i128 = __int128;
using u128 = unsigned __int128;

随机数生成

// std::mt19937
mt19937 rng(std::chrono::steady_clock::now()
                .time_since_epoch()
                .count());
// splitmix64
struct custom_hash {
    static auto splitmix64(u64 x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    auto operator()(u64 x) const {
        static const u64 SEED =
            std::chrono::high_resolution_clock::
                now()
                    .time_since_epoch()
                    .count();
        return splitmix64(x + SEED);
    }
};

二分答案

int l, r, ans;
bool check(int);
while (l <= r) {
    int mid = l + (r - l) / 2;
    if (check(mid)) {
        ans = mid;
        // l 和 r 取决于方向
        l = mid + 1;
    } else {
        r = mid - 1;
    }
}

Lambda 递归

auto dfs1 = [&](auto&& self, int x,
                int p) -> void {
    for (auto y : adj[x]) {
        if (y == p)
            continue;
        self(self, y, x);
    }
};
dfs1(dfs1, 0, 0);
// C++17
std::function<void(int, int)> dfs2 =
    [&](int x, int p) {
        for (auto y : adj[x]) {
            if (y == p)
                continue;
            dfs2(y, x);
        }
    };

数学

线性筛

初始化 [2, N] 内质数和非质数标记。

template <int N>
struct Sieve {
    std::vector<int> pri;
    bool np[N + 1]{};

    Sieve() {
        np[0] = np[1] = true;
        for (int i = 2; i <= N; i++) {
            if (!np[i])
                pri.emplace_back(i);
            for (int p : pri) {
                if (i * p > N)
                    break;
                np[i * p] = true;
                if (i % p == 0)
                    break;
            }
        }
    }
};

因数/约数分解

Factor(n) 返回质因数分解，Divisor(n) 返回大于 1 且小于等于 n 的约数。

inline auto Factor(int n) {
    std::vector<std::pair<int, int>> res;
    for (int i = 2; i * i <= n; i++) {
        int cur = 0;
        while (n % i == 0) {
            cur++;
            n /= i;
        }
        if (cur != 0)
            res.emplace_back(i, cur);
    }
    if (n > 1)
        res.emplace_back(n, 1);
    return res;
}

inline auto Divisor(int n) {
    std::vector<int> res;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            res.emplace_back(i);
            if (i * i != n)
                res.emplace_back(n / i);
        }
    }
    std::sort(res.begin(), res.end());
    return res;
}

扩展欧几里得

返回 gcd(a, b)，并求出 ax + by = gcd(a, b) 的一组解。

template <class T>
inline T ExGCD(T a, T b, T& x, T& y) {
    if (b == 0) {
        x = 1, y = 0;
        return a;
    }
    auto d = ExGCD(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

快速幂

计算 a^b mod p。

inline auto PowMod(i64 a, i64 b, int p) {
    i64 res = 1;
    while (b) {
        if (b & 1)
            res = res * a % p;
        b = b >> 1;
        a = a * a % p;
    }
    return res;
}

组合数

基于 ModInt 预处理阶乘和逆阶乘。

template <class Z>
struct Comb {
    std::vector<Z> fac, ifac;

    Comb(int n) : fac(n + 1), ifac(n + 1) {
        fac[0] = 1;
        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i;
        ifac[n] = Z(1) / fac[n];
        for (int i = n; i >= 1; i--)
            ifac[i - 1] = ifac[i] * i;
    }

    auto C(int n, int k) const {
        if (k < 0 or k > n)
            return Z{};
        return fac[n] * ifac[k] * ifac[n - k];
    }

    auto A(int n, int k) const {
        if (k < 0 or k > n)
            return Z{};
        return fac[n] * ifac[n - k];
    }
};

线性基

维护异或线性空间。

using u64 = uint64_t;

template <class T = u64, int LOG = 63>
struct LinearBasis {
    std::array<T, LOG + 1> p{};
    int rank = 0;

    bool Insert(T x) {
        for (int i = LOG; i >= 0; i--) {
            if (((x >> i) & 1) == 0)
                continue;
            if (!p[i]) {
                p[i] = x;
                rank++;
                return true;
            }
            x ^= p[i];
        }
        return false;
    }

    bool Contains(T x) const {
        for (int i = LOG; i >= 0; i--) {
            if (((x >> i) & 1) == 0)
                continue;
            if (!p[i])
                return false;
            x ^= p[i];
        }
        return true;
    }

    T MaxXor(T x = 0) const {
        for (int i = LOG; i >= 0; i--)
            if ((x ^ p[i]) > x)
                x ^= p[i];
        return x;
    }

    std::vector<T> Basis() const {
        std::vector<T> res;
        for (int i = 0; i <= LOG; i++)
            if (p[i])
                res.emplace_back(p[i]);
        return res;
    }
};

数据结构

并查集 (DSU)

维护无向连通性和连通块大小，Merge(x, y) 返回是否真的合并成功。

struct DSU {
    std::vector<int> f, sz;

    DSU(int n) : f(n), sz(n, 1) {
        std::iota(f.begin(), f.end(), 0);
    }

    auto Find(int x) {
        while (x != f[x])
            x = f[x] = f[f[x]];
        return x;
    }

    auto Merge(int x, int y) {
        x = Find(x), y = Find(y);
        if (x == y)
            return false;
        if (sz[x] < sz[y])
            std::swap(x, y);
        sz[x] += sz[y];
        f[y] = x;
        return true;
    }

    auto Size(int x) { return sz[Find(x)]; }
};

树状数组

维护单点加、前缀和、区间和。

template <typename T>
struct Fenwick {
    int n;
    std::vector<T> a;

    Fenwick(int n) : n(n), a(n + 1) {}

    void Add(int x, T v) {
        for (; x <= n; x += x & -x)
            a[x] += v;
    }

    auto sum(int x) {
        T res = {};
        for (; x; x -= x & -x)
            res += a[x];
        return res;
    }

    auto Sum(int l, int r) {
        return sum(r) - sum(l - 1);
    }
};

线段树

维护单点修改、区间查询。

关于 S 的约束：

关于初始区间：

template <class S>
struct SegTree {

    int n;
    std::vector<S> tr;

    SegTree(int n, const S& e) {
        build(n, std::vector<S>(n, e));
    }

    template <class It>
    SegTree(It l, It r) {
        build(r - l, l);
    }

    template <class Arr>
    void build(int m, const Arr& arr) {
        for (n = 1; n < m; n <<= 1)
            ;
        tr.resize(n << 1);
        for (int i = 0; i < m; i++)
            tr[i + n] = arr[i];
        for (int i = n - 1; i >= 1; i--)
            pull(i);
    }

    void pull(int k) {
        tr[k] = tr[k << 1] + tr[k << 1 | 1];
    }

    void Set(int p, const S& x) {
        p += n;
        tr[p] = x;
        for (p >>= 1; p; p >>= 1)
            pull(p);
    }

    auto Get(int p) { return tr[p + n]; }

    auto Query(int l, int r) {
        l += n, r += n;
        S sml{}, smr{};
        while (l < r) {
            if (l & 1)
                sml = sml + tr[l++];
            if (r & 1)
                smr = tr[--r] + smr;
            l >>= 1;
            r >>= 1;
        }
        return sml + smr;
    }
};

懒标记线段树

使用非递归的实现方式。

关于 F 的约束：

关于 S 的约束：

关于初始区间：

template <class S, class F>
struct LazySegTree {

    int n, h;
    std::vector<S> tr;
    std::vector<F> lz;

    LazySegTree(int n, const S& e) {
        build(n, std::vector<S>(n, e));
    }

    template <class It>
    LazySegTree(It l, It r) {
        build(r - l, l);
    }

    template <class Arr>
    void build(int m, const Arr& arr) {
        for (n = 1; n < m; n <<= 1)
            ;
        h = __builtin_ctz(n);
        tr.resize(n << 1);
        lz.resize(n);
        for (int i = 0; i < m; i++)
            tr[i + n] = arr[i];
        for (int i = n - 1; i >= 1; i--)
            pull(i);
    }

    void apply(int k, const F& f) {
        tr[k] *= f;
        if (k < n)
            lz[k] += f;
    }

    void pull(int k) {
        tr[k] = tr[k << 1] + tr[k << 1 | 1];
    }

    void push(int k) {
        apply(k << 1, lz[k]);
        apply(k << 1 | 1, lz[k]);
        lz[k] = {};
    }

    void Set(int p, const S& x) {
        p += n;
        for (int i = h; i >= 1; i--)
            push(p >> i);
        tr[p] = x;
        for (int i = 1; i <= h; i++)
            pull(p >> i);
    }

    auto Get(int p) {
        p += n;
        for (int i = h; i >= 1; i--)
            push(p >> i);
        return tr[p];
    }

    void Update(int l, int r, const F& f) {
        l += n, r += n;
        for (int i = h; i >= 1; i--) {
            if ((l & ((1 << i) - 1)) != 0)
                push(l >> i);
            if ((r & ((1 << i) - 1)) != 0)
                push((r - 1) >> i);
        }
        {
            int l_ = l, r_ = r;
            while (l < r) {
                if (l & 1)
                    apply(l++, f);
                if (r & 1)
                    apply(--r, f);
                l >>= 1;
                r >>= 1;
            }
            l = l_;
            r = r_;
        }
        for (int i = 1; i <= h; i++) {
            if ((l & ((1 << i) - 1)) != 0)
                pull(l >> i);
            if ((r & ((1 << i) - 1)) != 0)
                pull((r - 1) >> i);
        }
    }

    auto Query(int l, int r) {
        l += n, r += n;
        for (int i = h; i >= 1; i--) {
            if ((l & ((1 << i) - 1)) != 0)
                push(l >> i);
            if ((r & ((1 << i) - 1)) != 0)
                push((r - 1) >> i);
        }
        S sml{}, smr{};
        while (l < r) {
            if (l & 1)
                sml = sml + tr[l++];
            if (r & 1)
                smr = tr[--r] + smr;
            l >>= 1;
            r >>= 1;
        }
        return sml + smr;
    }
};

异或 Trie

维护非负整数可重集，支持插入、删除和查询与 x 异或的最大值。

struct XorTrie {
    std::vector<std::array<int, 2>> ch;
    std::vector<int> cnt;
    int tot = 1;

    XorTrie(int n) : ch(n * 32), cnt(n * 32) {}

    void Insert(int x) {
        int u = 1;
        cnt[u]++;
        for (int i = 30; i >= 0; i--) {
            int c = (x >> i) & 1;
            if (not ch[u][c]) {
                ch[u][c] = ++tot;
            }
            u = ch[u][c];
            cnt[u]++;
        }
    }

    void Erase(int x) {
        int u = 1;
        cnt[u]--;
        for (int i = 30; i >= 0; i--) {
            int c = (x >> i) & 1;
            u = ch[u][c];
            cnt[u]--;
        }
    }

    auto Query(int x) {
        int res = 0;
        int u = 1;
        for (int i = 30; i >= 0; i--) {
            int c = (x >> i) & 1;
            int v = ch[u][c ^ 1];
            if (v and cnt[v] > 0) {
                u = v;
                res |= (1 << i);
            } else {
                u = ch[u][c];
            }
        }
        return res;
    }
};

ModInt

固定模数整数，支持四则运算、输入输出和比较。

using u32 = unsigned;
using i64 = int64_t;
using u64 = uint64_t;

template <class T>
constexpr auto power(T a, u64 b, T res = 1) {
    for (; b != 0; b /= 2, a *= a) {
        if (b & 1) {
            res *= a;
        }
    }
    return res;
}

template <u32 P>
constexpr auto mulMod(u32 a, u32 b) {
    return u32(u64(a) * b % P);
}

template <u64 P>
constexpr auto mulMod(u64 a, u64 b) {
    u64 res =
        a * b - u64(1.L * a * b / P - 0.5L) * P;
    res %= P;
    return res;
}

constexpr auto safeMod(i64 x, i64 m) {
    x %= m;
    if (x < 0) {
        x += m;
    }
    return x;
}

constexpr std::pair<i64, i64> invGcd(i64 a, i64 b) {
    a = safeMod(a, b);
    if (a == 0) {
        return {b, 0};
    }

    i64 s = b, t = a;
    i64 m0 = 0, m1 = 1;

    while (t) {
        i64 u = s / t;
        s -= t * u;
        m0 -= m1 * u;

        std::swap(s, t);
        std::swap(m0, m1);
    }

    if (m0 < 0) {
        m0 += b / s;
    }

    return {s, m0};
}

template <class U, U P>
struct ModIntBase {
    static_assert(std::is_unsigned<U>::value);

    constexpr ModIntBase() : x(0) {}
    template <class T, std::enable_if_t<std::is_unsigned<T>::value,
                                       int> = 0>
    constexpr ModIntBase(T x_) : x(x_ % mod()) {}
    template <class T, std::enable_if_t<std::is_signed<T>::value,
                                       int> = 0>
    constexpr ModIntBase(T x_) {
        using S = std::make_signed_t<U>;
        S v = x_ % S(mod());
        if (v < 0) {
            v += mod();
        }
        x = v;
    }

    constexpr static auto mod() { return P; }

    constexpr auto val() const { return x; }

    constexpr auto operator-() const {
        ModIntBase res;
        res.x = (x == 0 ? 0 : mod() - x);
        return res;
    }

    constexpr auto inv() const {
        return power(*this, mod() - 2);
    }

    constexpr auto&
    operator*=(const ModIntBase& rhs) & {
        x = mulMod<mod()>(x, rhs.val());
        return *this;
    }
    constexpr auto&
    operator+=(const ModIntBase& rhs) & {
        x += rhs.val();
        if (x >= mod()) {
            x -= mod();
        }
        return *this;
    }
    constexpr auto&
    operator-=(const ModIntBase& rhs) & {
        x -= rhs.val();
        if (x >= mod()) {
            x += mod();
        }
        return *this;
    }
    constexpr auto&
    operator/=(const ModIntBase& rhs) & {
        return *this *= rhs.inv();
    }

    friend constexpr auto
    operator*(ModIntBase lhs,
              const ModIntBase& rhs) {
        lhs *= rhs;
        return lhs;
    }
    friend constexpr auto
    operator+(ModIntBase lhs,
              const ModIntBase& rhs) {
        lhs += rhs;
        return lhs;
    }
    friend constexpr auto
    operator-(ModIntBase lhs,
              const ModIntBase& rhs) {
        lhs -= rhs;
        return lhs;
    }
    friend constexpr auto
    operator/(ModIntBase lhs,
              const ModIntBase& rhs) {
        lhs /= rhs;
        return lhs;
    }

    friend constexpr auto&
    operator>>(std::istream& is, ModIntBase& a) {
        i64 i;
        is >> i;
        a = i;
        return is;
    }
    friend constexpr auto&
    operator<<(std::ostream& os,
               const ModIntBase& a) {
        return os << a.val();
    }

    friend constexpr bool
    operator==(const ModIntBase& lhs,
               const ModIntBase& rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool
    operator!=(const ModIntBase& lhs,
               const ModIntBase& rhs) {
        return !(lhs == rhs);
    }
    friend constexpr bool
    operator<(const ModIntBase& lhs,
              const ModIntBase& rhs) {
        return lhs.val() < rhs.val();
    }
    friend constexpr bool
    operator>(const ModIntBase& lhs,
              const ModIntBase& rhs) {
        return rhs < lhs;
    }
    friend constexpr bool
    operator<=(const ModIntBase& lhs,
               const ModIntBase& rhs) {
        return !(rhs < lhs);
    }
    friend constexpr bool
    operator>=(const ModIntBase& lhs,
               const ModIntBase& rhs) {
        return !(lhs < rhs);
    }

    U x;
};

template <u32 P>
using ModInt = ModIntBase<u32, P>;
template <u64 P>
using ModInt64 = ModIntBase<u64, P>;

#pragma once

#include #include #include #include

ST 表

维护静态区间查询。

template <class T>
struct SparseTable {
    using Op = std::function<T(const T&, const T&)>;

    int n = 0;
    Op op;
    std::vector<int> lg;
    std::vector<std::vector<T>> st;

    SparseTable() = default;

    SparseTable(const std::vector<T>& a, Op op)
        : op(std::move(op)) {
        Build(a.begin(), a.end());
    }

    template <class It>
    SparseTable(It l, It r, Op op)
        : op(std::move(op)) {
        Build(l, r);
    }

    template <class It>
    void Build(It l, It r) {
        n = int(r - l);
        lg.assign(n + 1, 0);
        for (int i = 2; i <= n; i++)
            lg[i] = lg[i >> 1] + 1;
        st.assign(lg[n] + 1, std::vector<T>(n));
        for (int i = 0; i < n; i++)
            st[0][i] = *(l + i);
        for (int k = 1; k < int(st.size()); k++) {
            int len = 1 << k;
            for (int i = 0; i + len <= n; i++) {
                st[k][i] =
                    op(st[k - 1][i],
                       st[k - 1][i + (len >> 1)]);
            }
        }
    }

    T Query(int l, int r) const {
        assert(0 <= l and l < r and r <= n);
        int k = lg[r - l];
        return op(st[k][l], st[k][r - (1 << k)]);
    }
};

图论

单源最短路

适用于非负边权图。

template <class Adj>
auto dijkstra(const Adj& adj, int n, int s) {
    vector<int> dis(n + 1, INF);
    vector<bool> vis(n + 1, false);
    priority_queue<pair<int, int>> pq;
    dis[s] = 0;
    pq.emplace(0, s);
    while (!pq.empty()) {
        auto [_, u] = pq.top();
        pq.pop();
        if (vis[u])
            continue;
        vis[u] = true;
        for (auto [v, w] : adj[u]) {
            if (dis[u] + w < dis[v]) {
                dis[v] = dis[u] + w;
                pq.emplace(-dis[v], v);
            }
        }
    }
    return dis;
}

分层图最短路

适用于非负边权图上最多使用 k 次特殊操作。

P4568 的建模方式：

复杂度 O((n k + m k) log(n k))。

template <class Adj>
auto LayeredDijkstra(const Adj& adj, int n, int s,
                     int k) {
    vector dist(n + 1, vector<i64>(k + 1, INF));
    priority_queue<tuple<i64, int, int>> pq;
    dist[s][0] = 0;
    pq.emplace(0, s, 0);
    while (!pq.empty()) {
        auto [d, u, used] = pq.top();
        pq.pop();
        d = -d;
        if (d != dist[u][used])
            continue;
        for (auto [v, w] : adj[u]) {
            if (dist[u][used] + w < dist[v][used]) {
                dist[v][used] = dist[u][used] + w;
                pq.emplace(-dist[v][used], v, used);
            }
            if (used < k and
                dist[u][used] < dist[v][used + 1]) {
                dist[v][used + 1] = dist[u][used];
                pq.emplace(-dist[v][used + 1], v,
                           used + 1);
            }
        }
    }
    return dist;
}

强连通分量 (SCC)

Tarjan 求有向图强连通分量。

struct SCC {
    int n, now = 0, cnt = 0;
    vector<vector<int>> adj;
    vector<int> dfn, low, stk, ins, id;

    SCC(int n)
        : n(n), adj(n + 1), dfn(n + 1), low(n + 1),
          ins(n + 1), id(n + 1) {}

    void AddEdge(int u, int v) {
        adj[u].emplace_back(v);
    }

    void tarjan(int u) {
        dfn[u] = low[u] = ++now;
        stk.emplace_back(u);
        ins[u] = true;
        for (int v : adj[u]) {
            if (!dfn[v]) {
                tarjan(v);
                low[u] = min(low[u], low[v]);
            } else if (ins[v]) {
                low[u] = min(low[u], dfn[v]);
            }
        }
        if (dfn[u] == low[u]) {
            cnt++;
            while (true) {
                int x = stk.back();
                stk.pop_back();
                ins[x] = false;
                id[x] = cnt;
                if (x == u)
                    break;
            }
        }
    }

    std::pair<std::vector<int>, int> Work() {
        for (int i = 1; i <= n; i++)
            if (!dfn[i])
                tarjan(i);
        return {id, cnt};
    }
};

割点

Tarjan 求无向图割点。

auto CutVertex(const vector<vector<int>>& adj,
               int n) {
    vector<int> dfn(n + 1), low(n + 1), cut(n + 1),
        res;
    int now = 0;
    auto dfs = [&](auto&& self, int u,
                   int p) -> void {
        dfn[u] = low[u] = ++now;
        int child = 0;
        for (int v : adj[u]) {
            if (!dfn[v]) {
                child++;
                self(self, v, u);
                low[u] = min(low[u], low[v]);
                if (p != 0 and low[v] >= dfn[u])
                    cut[u] = true;
            } else if (v != p) {
                low[u] = min(low[u], dfn[v]);
            }
        }
        if (p == 0 and child >= 2)
            cut[u] = true;
    };
    for (int i = 1; i <= n; i++)
        if (!dfn[i])
            dfs(dfs, i, 0);
    for (int i = 1; i <= n; i++)
        if (cut[i])
            res.emplace_back(i);
    return res;
}

二分图最大匹配

匈牙利算法。

auto BipartiteMatching(
    const vector<vector<int>>& adj, int n, int m) {
    vector<int> mt(m + 1), vis(m + 1);
    int ans = 0, stamp = 0;
    auto dfs = [&](auto&& self,
                   int u) -> bool {
        for (int v : adj[u]) {
            if (vis[v] == stamp)
                continue;
            vis[v] = stamp;
            if (!mt[v] or self(self, mt[v])) {
                mt[v] = u;
                return true;
            }
        }
        return false;
    };
    for (int i = 1; i <= n; i++) {
        stamp++;
        ans += dfs(dfs, i);
    }
    return ans;
}

拓扑排序

Kahn 算法。

auto TopoSort(const vector<vector<int>>& adj,
              int n) {
    vector<int> indeg(n + 1), res;
    queue<int> q;
    for (int u = 1; u <= n; u++)
        for (int v : adj[u])
            indeg[v]++;
    for (int i = 1; i <= n; i++)
        if (!indeg[i])
            q.emplace(i);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        res.emplace_back(u);
        for (int v : adj[u])
            if (--indeg[v] == 0)
                q.emplace(v);
    }
    return res;
}

差分约束

Bellman-Ford 判负环并求一组可行解。

auto DifferenceConstraints(
    int n, const vector<array<int, 3>>& edges) {
    vector<i64> d(n + 1);
    for (int i = 1; i <= n; i++) {
        bool changed = false;
        for (auto [u, v, w] : edges) {
            if (d[v] > d[u] + w) {
                d[v] = d[u] + w;
                changed = true;
            }
        }
        if (!changed)
            break;
        if (i == n)
            return vector<i64>{};
    }
    return d;
}

LCA

倍增求树上最近公共祖先。

struct LCA {
    int n, LOG;
    vector<int> dep, siz;
    vector<vector<int>> up;

    LCA(const vector<vector<int>>& adj, int root = 1) {
        n = adj.size() - 1;
        LOG = __lg(n) + 1;
        dep.assign(n + 1, 0);
        siz.assign(n + 1, 1);
        up.assign(LOG, vector<int>(n + 1, root));

        auto dfs = [&](auto&& self, int u,
                       int p) -> void {
            up[0][u] = p;
            for (int i = 1; i < LOG; i++)
                up[i][u] = up[i - 1][up[i - 1][u]];
            for (int v : adj[u]) {
                if (v == p)
                    continue;
                dep[v] = dep[u] + 1;
                self(self, v, u);
                siz[u] += siz[v];
            }
        };
        dfs(dfs, root, root);
    }

    int Get(int u, int v) const {
        if (dep[u] < dep[v])
            swap(u, v);
        u = jump(u, dep[u] - dep[v]);
        if (u == v)
            return u;
        for (int i = LOG - 1; i >= 0; i--) {
            if (up[i][u] != up[i][v]) {
                u = up[i][u];
                v = up[i][v];
            }
        }
        return up[0][u];
    }

    int Dis(int u, int v) const {
        int g = Get(u, v);
        return dep[u] + dep[v] - 2 * dep[g];
    }

    int Kth(int u, int v, int k) const {
        int g = Get(u, v);
        int du = dep[u] - dep[g];
        int d = du + dep[v] - dep[g];
        if (k <= du)
            return jump(u, k);
        return jump(v, d - k);
    }

    int Component(int u, int v) const {
        if (up[0][v] == u)
            return siz[v];
        return n - siz[u];
    }

    int jump(int u, int k) const {
        for (int i = 0; i < LOG; i++)
            if (k >> i & 1)
                u = up[i][u];
        return u;
    }
};

树上差分

对树上路径做批量加法，再一次 DFS 汇总。

struct TreeDifference {
    int n, LOG;
    vector<vector<int>> adj, up;
    vector<int> dep;
    vector<i64> diff;

    TreeDifference(const vector<vector<int>>& adj,
                   int root = 1)
        : n(adj.size() - 1), adj(adj),
          LOG(__lg(n) + 1), up(LOG, vector<int>(n + 1, root)),
          dep(n + 1), diff(n + 1) {
        auto dfs = [&](auto&& self, int u,
                       int p) -> void {
            up[0][u] = p;
            for (int i = 1; i < LOG; i++)
                up[i][u] = up[i - 1][up[i - 1][u]];
            for (int v : adj[u]) {
                if (v == p)
                    continue;
                dep[v] = dep[u] + 1;
                self(self, v, u);
            }
        };
        dfs(dfs, root, root);
    }

    void AddVertexPath(int u, int v,
                       i64 w = 1) {
        int g = lca(u, v);
        diff[u] += w;
        diff[v] += w;
        diff[g] -= w;
        if (up[0][g] != g)
            diff[up[0][g]] -= w;
    }

    void AddEdgePath(int u, int v,
                     i64 w = 1) {
        int g = lca(u, v);
        diff[u] += w;
        diff[v] += w;
        diff[g] -= 2 * w;
    }

    vector<i64> Work(int root = 1) {
        auto res = diff;
        auto dfs = [&](auto&& self, int u,
                       int p) -> void {
            for (int v : adj[u]) {
                if (v == p)
                    continue;
                self(self, v, u);
                res[u] += res[v];
            }
        };
        dfs(dfs, root, root);
        return res;
    }

    int jump(int u, int k) const {
        for (int i = 0; i < LOG; i++)
            if (k >> i & 1)
                u = up[i][u];
        return u;
    }

    int lca(int u, int v) const {
        if (dep[u] < dep[v])
            swap(u, v);
        u = jump(u, dep[u] - dep[v]);
        if (u == v)
            return u;
        for (int i = LOG - 1; i >= 0; i--) {
            if (up[i][u] != up[i][v]) {
                u = up[i][u];
                v = up[i][v];
            }
        }
        return up[0][u];
    }
};

Dinic 最大流

分层图 + 当前弧优化求最大流。

struct Dinic {
    struct Edge {
        int to, rev;
        i64 cap;
    };

    int n;
    vector<vector<Edge>> adj;
    vector<int> dep, cur;

    Dinic(int n)
        : n(n), adj(n + 1), dep(n + 1), cur(n + 1) {}

    int AddEdge(int u, int v, i64 c) {
        int id = adj[u].size();
        int revU = adj[v].size();
        int revV = adj[u].size();
        Edge a{v, revU, c};
        Edge b{u, revV, 0};
        adj[u].emplace_back(a);
        adj[v].emplace_back(b);
        return id;
    }

    i64 Flow(int u, int id) const {
        const Edge& e = adj[u][id];
        return adj[e.to][e.rev].cap;
    }

    i64 MaxFlow(int s, int t) {
        i64 flow = 0;
        constexpr i64 INF = numeric_limits<i64>::max() / 4;
        while (bfs(s, t)) {
            fill(cur.begin(), cur.end(), 0);
            while (i64 f = dfs(s, t, INF))
                flow += f;
        }
        return flow;
    }

    bool bfs(int s, int t) {
        fill(dep.begin(), dep.end(), -1);
        queue<int> q;
        dep[s] = 0;
        q.emplace(s);
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (auto& e : adj[u]) {
                if (e.cap > 0 and dep[e.to] == -1) {
                    dep[e.to] = dep[u] + 1;
                    q.emplace(e.to);
                }
            }
        }
        return dep[t] != -1;
    }

    i64 dfs(int u, int t, i64 f) {
        if (u == t or f == 0)
            return f;
        for (int& i = cur[u]; i < (i64)adj[u].size(); i++) {
            Edge& e = adj[u][i];
            if (e.cap <= 0 or dep[e.to] != dep[u] + 1)
                continue;
            i64 w = dfs(e.to, t, min(f, e.cap));
            if (!w)
                continue;
            e.cap -= w;
            adj[e.to][e.rev].cap += w;
            return w;
        }
        return 0;
    }
};

Hopcroft-Karp

二分图最大匹配。

struct HopcroftKarp {
    int n, m;
    vector<vector<int>> adj;
    vector<int> matchL, matchR, dis;

    HopcroftKarp(int n, int m)
        : n(n), m(m), adj(n + 1), matchL(n + 1),
          matchR(m + 1), dis(n + 1) {}

    void AddEdge(int u, int v) {
        adj[u].emplace_back(v);
    }

    int Work() {
        int ans = 0;
        while (bfs()) {
            for (int u = 1; u <= n; u++)
                if (!matchL[u] and dfs(u))
                    ans++;
        }
        return ans;
    }

    bool bfs() {
        queue<int> q;
        fill(dis.begin(), dis.end(), -1);
        for (int u = 1; u <= n; u++) {
            if (!matchL[u]) {
                dis[u] = 0;
                q.emplace(u);
            }
        }
        bool found = false;
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            for (int v : adj[u]) {
                int x = matchR[v];
                if (!x) {
                    found = true;
                } else if (dis[x] == -1) {
                    dis[x] = dis[u] + 1;
                    q.emplace(x);
                }
            }
        }
        return found;
    }

    bool dfs(int u) {
        for (int v : adj[u]) {
            int x = matchR[v];
            if (!x or (dis[x] == dis[u] + 1 and dfs(x))) {
                matchL[u] = v;
                matchR[v] = u;
                return true;
            }
        }
        dis[u] = -1;
        return false;
    }
};

字符串

KMP

求前缀函数，支持模式串匹配。

auto Kmp(const string& s) {
    int n = s.size();
    vector<int> f(n + 1);
    for (int i = 1, j = 0; i < n; i++) {
        while (j > 0 and s[i] != s[j])
            j = f[j];
        j += (s[i] == s[j]);
        f[i + 1] = j;
    }
    return f;
}

多项式

FFT

复数 FFT 求整数多项式卷积。

using i64 = int64_t;
using comp = complex<double>;
const double PI = acos(-1);

void Fft(vector<comp>& a, bool inv) {
    int n = a.size();
    for (int i = 1, j = 0; i < n; i++) {
        int bit = n >> 1;
        for (; j & bit; bit >>= 1)
            j ^= bit;
        j ^= bit;
        if (i < j)
            swap(a[i], a[j]);
    }
    for (int len = 2; len <= n; len <<= 1) {
        double ang = 2 * PI / len * (inv ? -1 : 1);
        comp wlen(cos(ang), sin(ang));
        for (int i = 0; i < n; i += len) {
            comp w(1);
            for (int j = 0; j < len / 2; j++) {
                comp u = a[i + j];
                comp v = a[i + j + len / 2] * w;
                a[i + j] = u + v;
                a[i + j + len / 2] = u - v;
                w *= wlen;
            }
        }
    }
    if (inv)
        for (auto& x : a)
            x /= n;
}

auto Convolution(const vector<i64>& a,
                 const vector<i64>& b) {
    if (a.empty() or b.empty())
        return vector<i64>{};
    int need = a.size() + b.size() - 1;
    int n = 1;
    while (n < need)
        n <<= 1;
    vector<comp> fa(a.begin(), a.end()),
        fb(b.begin(), b.end());
    fa.resize(n);
    fb.resize(n);
    Fft(fa, false);
    Fft(fb, false);
    for (int i = 0; i < n; i++)
        fa[i] *= fb[i];
    Fft(fa, true);
    vector<i64> res(need);
    for (int i = 0; i < need; i++)
        res[i] = llround(fa[i].real());
    return res;
}