P4396 [AHOI2013] 作业 题解

题目链接：作业

其实可以类似“HH的项链”转化为偏序问题再 cdq 分治。不过这题感觉莫队然后值域分块很好写啊，基本不用动脑。

考虑查询的两个信息：

1. \(a \le x\le b\) 的 \(x\) 的个数。

2. \(a \le x \le b\) 的 \(x\) 的种类。

考虑优化莫队插入和查询复杂度，我们用值域分块。

值域分块常常有两种常见的写法：

1. 修改：\(O(1)\)，查询：\(\sqrt{V_{max}}\)。做法：信息维护每个块的信息，然后修改只需要修改特定的值域块信息，查询类似常规分块，同块暴力，不同块散块暴力，整块直接查询。

2. 修改：\(O(\sqrt{V_{max}})\)，查询：\(O(1)\)。做法：维护前缀值域块和块内前缀信息，修改直接修改前缀信息。查询使用前缀和作差去查。

这两种都常常用于平滑修改和查询的复杂度。

本题中，使用莫队查询，修改显然需要 \(O(1)\)，我们使用第一种即可。维护一个桶表示每种数的数量，每个值域块维护数量和种类两种信息，查询的时候就是常规分块查询了。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)

#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'
//用于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;
}

constexpr int N = 1e5 + 10;
int ansCnt[N], ansClass[N]; //两种答案
int pos[N]; //序列分块

struct Mo
{
    int l, r, id, a, b;

    bool operator<(const Mo& other) const
    {
        return pos[l] ^ pos[other.l] ? l < other.l : pos[l] & 1 ? r < other.r : r > other.r;
    }
} node[N];

//值域分块
constexpr int MX = 1e5;
constexpr int SIZE = sqrt(MX);
constexpr int CNT = (N + SIZE - 1) / SIZE;
int s[N], e[N], valPos[N]; //值域块的起点和终点、每个点对应值域块的编号
int cnt[N], valCnt[N], valClass[N]; //桶记录数量、值域块内数量、值域块内种类

inline void add(const int val)
{
    int curr = valPos[val];
    valCnt[curr]++;
    if (cnt[val]++ == 0)valClass[curr]++;
}

inline void del(const int val)
{
    int curr = valPos[val];
    valCnt[curr]--;
    if (--cnt[val] == 0)valClass[curr]--;
}

inline pii query(const int a, const int b)
{
    const int currL = valPos[a], currR = valPos[b];
    int ansCnt = 0, ansClass = 0;
    if (currL == currR)
    {
        forn(i, a, b)ansCnt += cnt[i], ansClass += cnt[i] != 0;
        return pii(ansCnt, ansClass);
    }
    forn(i, a, e[currL])ansCnt += cnt[i], ansClass += cnt[i] != 0;
    forn(i, s[currR], b)ansCnt += cnt[i], ansClass += cnt[i] != 0;
    forn(i, currL+1, currR-1)ansCnt += valCnt[i], ansClass += valClass[i];
    return pii(ansCnt, ansClass);
}

int n, q;
int val[N];

inline void solve()
{
    cin >> n >> q;
    int siz = sqrt(n);
    forn(i, 1, n)cin >> val[i], pos[i] = (i - 1) / siz + 1;
    forn(i, 1, q)
    {
        auto& [l,r,id,a,b] = node[i];
        cin >> l >> r >> a >> b, id = i;
    }
    forn(i, 1, MX)valPos[i] = (i - 1) / SIZE + 1;
    forn(i, 1, CNT)s[i] = (i - 1) * SIZE + 1, e[i] = i * SIZE;
    e[CNT] = MX;
    int l = 1, r = 0;
    sortArr(node, q);
    forn(i, 1, q)
    {
        auto [L,R,id,a,b] = node[i];
        while (l > L)add(val[--l]);
        while (l < L)del(val[l++]);
        while (r < R)add(val[++r]);
        while (r > R)del(val[r--]);
        auto [fst, snd] = query(a, b);
        ansCnt[id] = fst, ansClass[id] = snd;
    }
    forn(i, 1, q)cout << ansCnt[i] << " " << ansClass[i] << 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;
}

\[时间复杂度为 \ O(m\times (\sqrt{n}+\sqrt{V_{max}}))
\]