CF940F Machine Learning题解

合集 - codeforces题解集1(37)

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题目链接：洛谷 或者 CF

不是特别难的题，抽象下题意就是算区间次数出现的次数 mex 和带单点修改。看到范围 $1e5$ 还带修改，传统的 mex 求法里貌似就莫队类算法好带修，考虑带修莫队。

然而涉及到 mex 问题，你可能不由自主地想到回滚莫队求 mex 只删不加的板子题：P4137 Rmq Problem / mex。虽然求 mex 有很多种做法，但这种单点带修似乎都不好整，真得要搓一个回滚莫队和带修莫队的鬼畜结合吗？注意到题意问的是次数的次数 mex，这个东西是具有上界的：

$$\mathrm{考}\mathrm{虑}\mathrm{次}\mathrm{数}\mathrm{的}\mathrm{次}\mathrm{数}\mathrm{集}\mathrm{合}=\{1,2,3,4,5....\}，\mathrm{那}\mathrm{么}\mathrm{对}\mathrm{应}\mathrm{的}\mathrm{需}\mathrm{要}\mathrm{的}\mathrm{元}\mathrm{素}\mathrm{个}\mathrm{数}\mathrm{为}\sum _{i=1}^{mex}i$$

$$={\displaystyle \frac{(mex+1)\ast mex}{2}}\le n\Rightarrow mex\le \sqrt{n}$$

所以这玩意我们完全可以暴力啊，$n$ 次暴力查询最多也是 $n\sqrt{n}$ 的数量级。

算法框架

考虑带修莫队，带修莫队只需要在原来离线的时候把修改也单独离线出来，并且记录每次查询对应的是第几次修改。比如第 $i$ 次查询当前区间情况更新完毕，我们考虑上一回更新完毕的查询对应的修改次数为 $t$ 次，那么我们就不断地进行时间轴上的递进或者回滚。

黑色为时间轴，假如我们上一回确定的访问为 $q2$ ，当前准备确定 $q1$ 的答案，并且 $l$ 和 $r$ 已经移动到了正确位置。发现 $q1$ 处没有发生任何修改，而 $q2$ 处修改过 $1$ 次，所以我们考虑撤销修改，因为把修改离线了，所以很好地可以跑时间轴上的回撤修改，或者更新修改。具体的如果这个修改的地方会在 $[l,r]$ 上，则顺带更新一波答案，当做删除当前的，插入新的。更新完毕以后交换离线存的修改值和当前值。具体的，如果这是一波撤销操作，则离线存的就是原来的值，如果是更新操作，则是修改后的值。

另外带修莫队常见的会将块长设置为：$n^{\frac{2}{3}}$ 较优。注意离散化值域方便用数组作为桶计数。

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;
int pos[N];
//查询时间轴上的莫队
struct MoQuery
{
    int l, r, id, tim;
    //三个维度排序,奇偶块排序优化
    bool operator<(const MoQuery& other) const
    {
        if (pos[l] != pos[other.l])return pos[l] < pos[other.l];
        if (pos[r] != pos[other.r])return pos[r] < pos[other.r];
        return pos[r] & 1 ? tim > other.tim : tim < other.tim;
    }
} query[N];
//更新时间轴上的莫队
struct MoUpdate
{
    int pos, val;
} update[N];
 
int a[N], b[N];
int mx;
hash2<int, int> mp;
int cnt[N];
int cntCnt[N];
int cntQ, cntU;
 
inline void add(const int val)
{
    cntCnt[cnt[val]]--;
    cntCnt[++cnt[val]]++;
}
 
inline void del(const int val)
{
    cntCnt[cnt[val]]--;
    cntCnt[--cnt[val]]++;
}
 
//需要修改的在查询范围内就更新
inline void back(const int l, const int r, const int x)
{
    auto [pos,val] = update[x];
    if (l <= pos and pos <= r)
    {
        del(a[pos]);
        add(val);
    }
    swap(a[pos], update[x].val);
}
 
int ans[N];
int n, q;
int siz;
int op;
 
inline void solve()
{
    cin >> n >> q;
    int B = floor(pow(n, 2.0 / 3.0));
    forn(i, 1, n)cin >> a[i], b[++siz] = a[i], pos[i] = (i - 1) / B + 1;
    //离线查询,分两类,查询/修改
    forn(i, 1, q)
    {
        cin >> op;
        if (op == 1)
        {
            auto& [l,r,id,t] = query[++cntQ];
            cin >> l >> r;
            id = cntQ;
            t = cntU;//记录当前修改次数,即修改时间轴上的坐标
        }
        else
        {
            auto& [pos,val] = update[++cntU];
            cin >> pos >> val;
            b[++siz] = val;
        }
    }
    //离散化
    sortArr(b, siz);
    mx = disc(b, siz);
    forn(i, 1, mx)mp[b[i]] = i;
    forn(i, 1, n)a[i] = mp[a[i]];
    forn(i, 1, cntU)update[i].val = mp[update[i].val];
    sortArr(query, cntQ);
    int l = 1, r = 0, t = 0;
    forn(i, 1, cntQ)
    {
        auto [L,R,id,T] = query[i];
        while (l < L)del(a[l++]);
        while (r > R)del(a[r--]);
        while (l > L)add(a[--l]);
        while (r < R)add(a[++r]);
        //修改时间轴上的回撤和更新
        while (t < T)back(L, R, ++t);
        while (t > T)back(L, R, t--);
        int res = 1;
        //暴力单次不超过sqrt(n)
        while (cntCnt[res])res++;
        ans[id] = res;
    }
    forn(i, 1, cntQ)cout << ans[i] << 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();
}
 

$$\text{最终复杂度为暴力查询加修改 }O(m\sqrt{n}+m{n}^{\frac{2}{3}})$$