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P4113 [HEOI2012] 采花 题解

时间：2024-02-14 21:44:06浏览次数：22  
标签：typedef ch P4113 int 题解 HEOI2012 template include define

题目链接：采花

这题数据加强到卡了 \(2e6\) 的可持久化线段树在线做法，先给只 tle 了最后一个点的代码：

卡常参照代码 
#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;
}

namespace fasI
{
    constexpr int BF_SIZE = 1 << 12;
    bool IOerr = false;

    inline char nc()
    {
        static char buf[BF_SIZE], *p1 = buf + BF_SIZE, *pend = buf + BF_SIZE;
        if (p1 == pend)
        {
            p1 = buf;
            pend = buf + fread(buf, 1, BF_SIZE,stdin);
            if (pend == p1)
            {
                IOerr = true;
                return -1;
            }
        }
        return *p1++;
    }

    inline bool bla(const char ch) { return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t'; }

    template <typename T>
    void Rd(T& x)
    {
        char ch;
        while (bla(ch = nc()));
        T sign = 1;
        if (ch == '-')sign = -1, ch = nc();
        for (x = ch - '0'; (ch = nc()) >= '0' && ch <= '9'; x = x * 10 + ch - '0');
        x *= sign;
    }
#undef BF_SIZE
}

using namespace fasI;
constexpr int N = 2e6 + 10;

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

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

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

inline int query(const int L, const int R, const int pos, const int l = 1, const int r = n)
{
    if (!L and !R)return 0;
    if (l == r)return cnt(R) - cnt(L);
    const int mid = l + r >> 1;
    int rightCnt = cnt(right(R)) - cnt(right(L));
    if (pos <= mid)return rightCnt + query(left(L),left(R), pos, l, mid);
    return query(right(L),right(R), pos, mid + 1, r);
}

int pre[N], pre_pre[N];
int x;

inline void solve()
{
    Rd(n), Rd(c), Rd(m);
    forn(i, 1, n)
    {
        Rd(x);
        root[i] = root[i - 1];
        if (pre[x])add(root[i], root[i], pre[x], -1);
        if (pre_pre[x])add(root[i], root[i], pre[x] = pre_pre[x], 1);
        pre_pre[x] = i;
    }
    while (m--)
    {
        int l, r;
        Rd(l), Rd(r);
        write(endl, query(root[l - 1], root[r], l));
    }
}

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;
}

这玩意挺经典的，难怪这题给离线，题目翻译下，就是问区间上有多少种数的数量 \(\ge 2\)，首先在线可持久化线段树维护一下前驱贡献类似 HH的项链。不过卡常了，因为涉及到删除，所以预处理应该是 \(2\) 倍原先建树复杂度。

讲讲离线做法，扫描线处理下查询，从左往右修改，每加入一个数，如果原先有前，我们去掉前驱的贡献，并且我们开一个桶，当然这里由于只涉及到去区间上只有两个数，其实我们再记录前驱的前驱就能表示出上一个贡献位置，如果这题改为区间内该数的数量 \(\ge k\)，可以直接开桶找前面第 \(k\) 个位置即为贡献位置。将贡献位置平移一位，用新的前第 \(2\) 个数代替原有的贡献，树状数组上最终查询即可，读入比较大，可以考虑快读或者究极快读。

参照代码 
#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 = 2e6 + 10;
int bit[N];
int n, c, m;

inline void add(int x, const int val)
{
    while (x <= n)bit[x] += val, x += lowBit(x);
}

inline int query(int x)
{
    int ans = 0;
    while (x)ans += bit[x], x -= lowBit(x);
    return ans;
}

vector<pii> seg[N];
int pre[N], pre_pre[N];
int a[N];
int ans[N];

inline void solve()
{
    read(n, c, m);
    forn(i, 1, n)read(a[i]);
    forn(i, 1, m)
    {
        int l, r;
        read(l, r);
        seg[r].emplace_back(l, i);
    }
    forn(r, 1, n)
    {
        if (pre[a[r]])add(pre[a[r]], -1);
        if (pre_pre[a[r]])pre[a[r]] = pre_pre[a[r]], add(pre[a[r]], 1);
        pre_pre[a[r]] = r;
        for (const auto [l,id] : seg[r])ans[id] = query(r) - query(l - 1);
    }
    forn(i, 1, m)write(endl, ans[i]);
}

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;
}

\[时间复杂度为:\ O((n+m)\log{n}) \]

标签：typedef,ch,P4113,int,题解,HEOI2012,template,include,define
From： https://www.cnblogs.com/Athanasy/p/18015652

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