結果

問題 No.2342 Triple Tree Query (Hard)
ユーザー robinyqc
提出日時 2024-11-28 15:31:01
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3,226 ms / 10,000 ms
コード長 6,902 bytes
コンパイル時間 1,665 ms
コンパイル使用メモリ 101,524 KB
最終ジャッジ日時 2025-02-25 06:21:48
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 2
other AC * 36
権限があれば一括ダウンロードができます

ソースコード

diff #

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <tuple>
using namespace std;

using i32 = int;
using i64 = long long;
using u32 = unsigned int;
using u64 = unsigned long long;

template<typename T> using vec = vector<T>;

using pii = pair<u32, u32>;

const u32 P = 998244353;

struct mint
{
    u32 val;
    mint(): val() { }
    mint(u32 x): val(x) { }
    mint &operator+=(mint t) 
    {
        val += t.val;
        if (val >= P) val -= P;
        return *this;
    }
    mint &operator-=(mint t)
    {
        if (val < t.val) val += P;
        val -= t.val;
        return *this;
    }
    mint &operator*=(mint t)
    {
        val = (u64)val * t.val % P;
        return *this;
    }
    friend mint operator+(mint a, mint b) { return a += b; }
    friend mint operator-(mint a, mint b) { return a -= b; }
    friend mint operator*(mint a, mint b) { return a *= b; }
    
    mint pow(u32 y)
    {
        u64 res = 1, x = val;
        for (; y; y >>= 1) {
            if (y & 1) res = res * x % P;
            x = x * x % P;
        }
        return mint(res);
    }
    
    mint inv() { return pow(P - 2); }
    
    friend bool operator==(mint a, mint b) { return a.val == b.val; }
};

bool pairle(const pii &x, const pii &y)
{
    return x.first <= y.first && x.second <= y.second;
}

pii pairmin(const pii &x, const pii &y)
{
    return {min(x.first, y.first), min(x.second, y.second)};
}
pii pairmax(const pii &x, const pii &y)
{
    return {max(x.first, y.first), max(x.second, y.second)};
}

template<typename M>
struct KDTree
{
    using A = typename M::S;

    u32 n;
    vec<u32> rnk;
    
    struct Node
    {
        pii sp, ep;
        Node(): sp(), ep() { }
        Node(const pii &a): sp(a), ep(a) { }
    };
    
    vec<A> t;
    vec<Node> d;
    
    KDTree(u32 _n, const vec<pii> &p): n(_n), rnk(n)
    {
        u32 m = 1;
        while (m < n) m = m * 2;
        t.resize(m * 2, M::un());
        d.resize(m * 2);
        vec<u32> idx(n);
        for (u32 i = 0; i < n; i++) idx[i] = i;
        build<0>(p, idx, 0, n, 1);
    }
    
    template<u32 D>
    void build(const vec<pii> &p, vec<u32> &idx, u32 l, u32 r, u32 x)
    {
        if (l + 1 == r) {
            u32 id = idx[l];
            d[x] = p[id];
            rnk[id] = l;
            return ;
        }
        u32 mid = (l + r) >> 1;
        nth_element(idx.data() + l, idx.data() + mid, idx.data() + r, 
            [&](u32 x, u32 y) {return std::get<D>(p[x]) < std::get<D>(p[y]);});
        build<!D>(p, idx, l, mid, x * 2);
        build<!D>(p, idx, mid, r, x * 2 + 1);
        d[x].sp = pairmin(d[x * 2].sp, d[x * 2 + 1].sp);
        d[x].ep = pairmax(d[x * 2].ep, d[x * 2 + 1].ep);
    }

    void push_down(u32 x)
    {
        if (t[x] != M::un()) {
            M::oop(t[x * 2], t[x]);
            M::oop(t[x * 2 + 1], t[x]);
            t[x] = M::un();
        }
    }
    
    void apply(const pii &st, const pii &ed, const A &a)
    {
        apply_rec(st, ed, a, 0, n, 1);
    }
    A get(u32 x) 
    {
        return get_rec(rnk[x], 0, n, 1);
    }
    
    void apply_rec(const pii &st, const pii &ed, const A &a, 
        u32 l, u32 r, u32 x)
    {
        if (pairle(st, d[x].sp) && pairle(d[x].ep, ed)) {
            M::oop(t[x], a);
            return ;
        }
        if (!pairle(st, d[x].ep) || !pairle(d[x].sp, ed)) return ;
        u32 mid = (l + r) >> 1;
        push_down(x);
        apply_rec(st, ed, a, l, mid, x * 2);
        apply_rec(st, ed, a, mid, r, x * 2 + 1);
    }
    
    A get_rec(u32 p, u32 l, u32 r, u32 x)
    {
        if (l + 1 == r) return t[x];
        u32 mid = (l + r) >> 1;
        push_down(x);
        if (p < mid) return get_rec(p, l, mid, x * 2);
        return get_rec(p, mid, r, x * 2 + 1);
    }
};

struct MonoidAffine
{
    using S = pair<mint, mint>;
    static void oop(S &a, const S &b)
    {
        a.first *= b.first;
        (a.second *= b.first) += b.second;
    }
    static S un() { return {1, 0}; }
};

signed main()
{
    // freopen("tour.in", "r", stdin);
    // freopen("tour.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    u32 n, q;
    cin >> n >> q;
    vec<vec<u32>> g(n);
    for (u32 i = 1, u, v; i < n; i++) {
        cin >> u >> v;
        --u, --v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
    }
    vec<mint> a(n);
    for (u32 i = 0, x; i < n; i++) {
        cin >> x;
        a[i] = x;
    }
    
    vec<u32> dfn(n), rnk(n), dep(n), fa(n), dfnr(n), top(n), son(n, -1);
    u32 dfnn = 0;
    
    [&, dfs = [&](auto &&f, u32 x) -> void
    {
        dfn[x] = 1;
        for (u32 v: g[x]) if (v != fa[x]) {
            fa[v] = x;
            dep[v] = dep[x] + 1;
            f(f, v);
            dfn[x] += dfn[v];
            if (son[x] == -1u || dfn[v] > dfn[son[x]]) son[x] = v;
        }
    }](){dfs(dfs, 0);}();

    [&, dfs = [&](auto &&f, u32 x) -> void
    {
        dfn[x] = dfnn;
        rnk[dfnn++] = x;
        if (son[x] != -1u) {
            top[son[x]] = top[x];
            f(f, son[x]);
        }
        for (u32 v: g[x]) if (v != fa[x] && v != son[x]) {
            top[v] = v;
            f(f, v);
        }
        dfnr[x] = dfnn - 1;
    }](){dfs(dfs, 0);}();
    
    vec<pii> initp(n);
    for (u32 i = 0; i < n; i++) initp[i] = {dfn[i], dep[i]};
    
    KDTree<MonoidAffine> t(n, initp);
    
    while (q--) {
        u32 op;
        cin >> op;
        if (op == 1) {
            u32 x;
            cin >> x;
            --x;
            auto xx = t.get(x);
            cout << (xx.first * a[x] + xx.second).val << '\n';
        }
        else if (op == 4) {
            u32 x, y, k, b;
            cin >> x >> y >> k >> b;
            --x, --y;
            pair<mint, mint> val(k, b);
            while (top[x] != top[y]) {
                if (dep[top[x]] < dep[top[y]]) swap(x, y);
                t.apply({dfn[top[x]], 0}, {dfn[x], n}, val);
                x = fa[top[x]];
            }
            if (dep[x] > dep[y]) swap(x, y);
            t.apply({dfn[x], 0}, {dfn[y], n}, val);
        }
        else if (op == 3) {
            u32 x, k, b;
            cin >> x >> k >> b;
            --x;
            t.apply({dfn[x], 0}, {dfnr[x], n}, {k, b});
        }
        else {
            u32 x, r, k, b;
            cin >> x >> r >> k >> b;
            --x;
            mint kk(mint(k).inv());
            pair<mint, mint> val(k, b), ival(kk, mint(0) - kk * b);
            t.apply({dfn[x], dep[x]}, {dfnr[x], dep[x] + r}, val);
            while (r-- && x != 0) {
                if (r >= 1 && k != 0) {
                    t.apply({dfn[x], dep[x]}, 
                            {dfnr[x], dep[x] + r - 1}, ival);
                }
                x = fa[x];
                t.apply({dfn[x], dep[x]}, {dfnr[x], dep[x] + r}, val);
            }
        }
    }
    return 0;
}

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