CF678F Lena and Queries题解

题目链接：CF 或者 洛谷

可以看到查询和插入就是李超线段树的基本操作，但在原有基础上多了一个删除操作，李超线段树不支持删除操作，但支持可撤销和可持久化，所以我们容易想到外层再套一个线段树分治即可。本题用可撤销就远远足够了，很好写。

具体的，我们读入所有操作，对于操作一，为当前下标线段读入它的 \(k\) 与 \(b\)，并且定义它的开始时间从此时此刻开始，到结束第 \(n\) 次操作的时刻。对于操作二删除，我们只需要更改删除的线段的结束时间为此时此刻的 \(i\) 时刻即可。第三个操作查询操作，只需要在待查询的队列当中插入一个 \((i,x)\) 表示第 \(i\) 时刻查询 \(\max{(kx+b)}\) 。当线段树分治到单点时只需要判断和队首时刻是否相同来决定是否查询即可。我们建立一棵时间轴线段树，为每个线段的开始到结束时刻插入它的编号，然后跑一遍线段树分治 \(+\) 可撤销的动态开点李超树即可。

细节

常见的李超树会直接插入 \((k,b)\) 的键值对，也可以添加需要添加的线段 \(id\)，这里为了减少空间常数，我们选择后一种。其次由于查询的 \(x\) 值域很大，我们可以考虑动态开点李超树来避免离散化之类的时空开销。最后，我们用栈只需要保存原来的李超树节点保存的线段 \(id\) 在线段树分治回溯的时候直接赋值即可。

参照代码

#include <bits/stdc++.h>

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

// #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 = 3e5 + 10;
constexpr ll INF = LLONG_MIN;//最小值
//线段数组
struct Seg
{
    int k,b;

    Seg(const int k, const int b)
        : k(k),
          b(b)
    {
    }

    Seg() = default;

    ll getY(const ll x) const
    {
        return k * x + b;
    }
} seg[N];

int cnt;

struct Node
{
    int left, right;
    int segId;
} node[N];

#define left(x) node[x].left
#define right(x) node[x].right
#define segId(x) node[x].segId
stack<pii> back;
int root;
//添加线段到李超树中
inline void add(int& curr, const int l, const int r, int id)
{
    if (!curr)curr = ++cnt;
    if (!segId(curr))
    {
        back.emplace(curr,segId(curr));
        segId(curr) = id;
        return;
    }
    const int mid = l + r >> 1;
    if (seg[id].getY(mid) > seg[segId(curr)].getY(mid))
    {
        back.emplace(curr,segId(curr));
        swap(segId(curr), id);
    }
    if (l == r)return;
    if (seg[id].getY(l) > seg[segId(curr)].getY(l))add(left(curr), l, mid, id);
    if (seg[id].getY(r) > seg[segId(curr)].getY(r))add(right(curr), mid + 1, r, id);
}
//时间线段树
struct Time
{
    vector<int> k_b;
} qu[N << 2];
//李超树的查询
inline ll query(const int curr, const int l, const int r, const int pos)
{
    if (!segId(curr))return INF;//无线段就无穷小
    ll ans = seg[segId(curr)].getY(pos);
    const int mid = l + r >> 1;
    if (l == r)return ans;
    if (pos <= mid)uMax(ans, query(left(curr), l, mid, pos));
    else uMax(ans, query(right(curr), mid + 1, r, pos));
    return ans;
}
//答案
vector<ll> ans;
//增加一条从l到r时时刻的时间线段
inline void addTimeSeg(const int curr, const int l, const int r, const int s, const int e, const int id)
{
    if (l <= s and e <= r)
    {
        qu[curr].k_b.push_back(id);
        return;
    }
    const int mid = s + e >> 1;
    if (l <= mid)addTimeSeg(ls(curr), l, r, s, mid, id);
    if (r > mid)addTimeSeg(rs(curr), l, r, mid + 1, e, id);
}

int l[N], r[N];
queue<pii> ask;//询问队列,(下标,查询的x)

//线段树分治
inline void dfs(const int curr, const int l, const int r)
{
    int pre = back.size();
    for (const auto id : qu[curr].k_b)add(root, -1e9, 1e9, id);
    const int mid = l + r >> 1;
    if (l == r)
    {
        if (!ask.empty() and ask.front().first == l)ans.push_back(query(root, -1e9, 1e9, ask.front().second)), ask.pop();
    }
    else dfs(ls(curr), l, mid), dfs(rs(curr), mid + 1, r);
    while (back.size() > pre)segId(back.top().first) = back.top().second, back.pop();//回溯撤销插入
}

int n;

inline void solve()
{
    cin >> n;
    forn(i, 1, n)
    {
        int op;
        cin >> op;
        if (op == 3)
        {
            int x;
            cin >> x;
            ask.emplace(i, x);//插入查询队列
        }
        else if (op == 1)
        {
            auto& [k,b] = seg[i];
            cin >> k >> b;
            l[i] = i, r[i] = n;//插入时间到结束时间
        }
        else
        {
            int pos;
            cin >> pos;
            r[pos] = i;//更改结束时间
        }
    }
    forn(i, 1, n)if (l[i])addTimeSeg(1, l[i], r[i], 1, n, i);//有线段就插入
    dfs(1, 1, n);//线段树分治
    for (const auto res : ans)
    {
        if (res == INF)cout << "EMPTY SET" << endl;
        else cout << res << endl;
    }
}

signed int main()
{
    Spider
    //------------------------------------------------------
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
}

\[最终时间复杂度由于李超树插入的是直线所以加上线段树分治应为\ O(nlog^{2}{n})
\]