P2633 Count on a tree 題解

Athanasy發表於2024-03-17

題目連結：Count on a tree

大概可以認為是樹上主席樹的板子

我在之前的某些題解提到了，主席樹一般來說有兩個基本功能：

  1. 可持久化功能，可以選擇回退或者新增版本。

  2. 對於可差性問題，可以有更好的轉化為字首和做法，常見的問題為權值型別問題。

在樹上的路徑第 \(k\) 大，顯然如果我們能拿到這條路徑上的權值樹顯然就隨便做二分了，很基礎的問題。

樹上字首和：

  1. 關於點的樹上字首和，我們對於一條路徑上的點假如設：\(pre[x]\) 為從 \(1 \sim x\) 的所有點，那麼有 \(u \rightarrow v\) 上的所有點可以抽象的表示為：\(pre[u]+pre[v]-pre[lca]-pre[fa[lca]]\)

  2. 關於邊的樹上字首和，\(pre[x]\) 表示 \(1 \sim x\) 的所有邊，那麼 \(u \rightarrow v\) 的所有邊可以表示為：\(pre[u]+pre[v]-2\times pre[lca]\)

如圖所示，這也是樹上字首和問題的常見兩種模型。本題顯然是常見的點模型，那麼樹上字首和就是一個很典型的差分型問題，那麼我們就可以透過四棵權值樹同時做貢獻即可拿到正確的權值樹。同時初始化，根據樹上字首和初始化規則，直接在 \(dfs\) 時我們做樹上字首和更新即可。最後，我們需要注意離散化值域以及求 \(lca\)，這裡就使用倍增求了。

參照程式碼 
#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

#define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用於Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

struct Hash
{
    static uint64_t splitmix64(uint64_t x)
    {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ x >> 30) * 0xbf58476d1ce4e5b9;
        x = (x ^ x >> 27) * 0x94d049bb133111eb;
        return x ^ x >> 31;
    }

    static size_t get(const uint64_t x)
    {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }

    template <typename T>
    size_t operator()(T x) const
    {
        return get(std::hash<T>()(x));
    }

    template <typename F, typename S>
    size_t operator()(pair<F, S> p) const
    {
        return get(std::hash<F>()(p.first)) ^ std::hash<S>()(p.second);
    }
};

constexpr int N = 1e5 + 10;
constexpr int T = ceil(log2(N));
int a[N], ord[N];
hash2<int, int, Hash> mp;
int fa[N][T + 1], deep[N];
vector<int> child[N];
int n, m, mx;

struct Node
{
    int left, right, cnt;
} node[N << 5];

int root[N];
#define left(x) node[x].left
#define right(x) node[x].right
#define cnt(x) node[x].cnt
int cnt;

inline void add(const int pre, int& curr, const int val, const int l = 1, const int r = mx)
{
    node[curr = ++cnt] = node[pre];
    cnt(curr)++;
    const int mid = l + r >> 1;
    if (l == r)return;
    if (val <= mid)add(left(pre),left(curr), val, l, mid);
    else add(right(pre),right(curr), val, mid + 1, r);
}

inline int query(const int u, const int v, const int lca, const int lca_fa, const int k, const int l = 1,
                 const int r = mx)
{
    if (l == r)return ord[l];
    const int leftSize = cnt(left(u)) + cnt(left(v)) - cnt(left(lca)) - cnt(left(lca_fa));
    const int mid = l + r >> 1;
    if (leftSize >= k)return query(left(u),left(v),left(lca),left(lca_fa), k, l, mid);
    return query(right(u),right(v),right(lca),right(lca_fa), k - leftSize, mid + 1, r);
}

inline void dfs(const int curr, const int pa)
{
    deep[curr] = deep[fa[curr][0] = pa] + 1;
    add(root[pa], root[curr], a[curr]);
    forn(i, 1, T)fa[curr][i] = fa[fa[curr][i - 1]][i - 1];
    for (const int nxt : child[curr])if (nxt != pa)dfs(nxt, curr);
}

inline int LCA(int x, int y)
{
    if (deep[x] < deep[y])swap(x, y);
    forv(i, T, 0)if (deep[fa[x][i]] >= deep[y])x = fa[x][i];
    if (x == y)return x;
    forv(i, T, 0)if (fa[x][i] != fa[y][i])x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

int last;

inline void solve()
{
    cin >> n >> m;
    forn(i, 1, n)cin >> a[i], ord[i] = a[i];
    sortArr(ord, n), mx = disc(ord, n);
    forn(i, 1, mx)mp[ord[i]] = i;
    forn(i, 1, n)a[i] = mp[a[i]];
    forn(i, 1, n-1)
    {
        int u, v;
        cin >> u >> v;
        child[u].push_back(v), child[v].push_back(u);
    }
    dfs(1, 0);
    while (m--)
    {
        int u, v, k;
        cin >> u >> v >> k, u ^= last;
        const int lca = LCA(u, v);
        const int lca_fa = fa[lca][0];
        cout << (last = query(root[u], root[v], root[lca], root[lca_fa], k)) << endl;
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[時間複雜度為:\ O((n+m)\log{n}) \]

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