CF1000F One Occurrence题解

合集 - codeforces题解集1(37)

1.CF1916E Happy Life in University 题解2023-12-312.CF763E Timofey and our friends animals题解2024-01-033.CF1270G Subset with Zero Sum2024-01-054.CF1045G AI robots题解2024-01-055.CF940F Machine Learning题解2024-01-066.CF678F Lena and Queries题解2024-01-127.CF1921F Sum of Progression 题解2024-01-168.CF526F Pudding Monsters 题解2024-01-219.CF455D Serega and Fun 题解2024-01-2210.CF351D Jeff and Removing Periods 题解2024-01-2211.CF452F Permutation 与 P2757 [国家集训队] 等差子序列 题解2024-01-2312.CF911G Mass Change Queries 题解2024-01-2413.CF145E Lucky Queries 题解2024-01-2514.CF1515F Phoenix and Earthquake 题解2024-01-2615.CF765F Souvenirs 题解2024-01-2716.CF1764H Doremy's Paint 2 题解2024-01-29

17.CF1000F One Occurrence题解2024-01-30

18.CF813E Army Creation 题解2024-01-3119.CF1706E Qpwoeirut and Vertices 题解2024-02-0120.CF620E New Year Tree 题解2024-02-0221.CF1454F Array Partition 题解2024-02-0322.CF1771F Hossam and Range Minimum Query 题解2024-02-0823.CF1931F Chat Screenshots 另一种题解2024-02-1424.CF896C Willem, Chtholly and Seniorious 题解2024-02-1525.CF55D Beautiful numbers 题解2024-02-1626.CF916E Jamie and Tree 题解2024-02-2327.CF696B Puzzles 题解2024-02-2828.CF383C Propagating tree 题解2024-02-2929.CF1436E Complicated Computations 题解2024-03-0830.CF817F MEX Queries 题解2024-03-2031.CF1638E Colorful Operations 题解2024-03-2132.CF1618G Trader Problem 题解2024-03-2233.CF794F Leha and security system 题解2024-03-2334.CF371E Subway Innovation 题解2024-04-0435.CF1200E Compress Words 题解2024-04-0636.CF1884D Counting Rhyme 题解2024-05-1737.CF1982F Sorting Problem Again 题解2024-06-26

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题目链接：CF 或者 洛谷

感觉很经典的题，而且给的 $5e5$，虽然莫队之类的很好想，但完全没必要去考虑这类算法，这种数据范围常数又大又开盲盒。很显然的具有单 $log$ 的算法。

回忆下经典问题 “HH的项链”。其实对于区间颜色数的方法网上已经总结的很全了，常见的无非就莫队，维护 $last$ 用 BIT、主席树之类的计数。这题和颜色数没啥太大关系，但分析方式是一致的。

考虑 $[l,r]$ 区间上的若干个数 $a_i$ 表示，如果有 $a_i$ 前面与自己相同的 $a_j$ 满足了 $j\ge l$，那么很显然这个区间 $a_i$ 至少出现了两次。考虑下所有数的 $pr{e}_{min}<l$，这显然是一个大前提。如果最小的 $pre$ 都不满足，那显然不行的。然后沿用 CF1916E Happy Life in University 题解。我们考虑使用后一个数代替前一个相同数的影响：

显然，前一个数 $x_1$ 的影响被后一个数 $x_2$ 代替了，那么 $x_1$ 不该对 $x$ 这个值产生任何影响，注意到查询为 $pre<l$，我们只需要把 $x_1$ 处的 $pre$ 设做 $INF$ 即可。接下来加入这个新的 $x_2$ 贡献代替原来的贡献，它的 $pre$ 显然为 $x_1$。然后在这个前提下用线段树维护单点覆盖和查询最小的 $pr$ 以及对应的值即可。查询用扫描线离线下来从左往右依次修改询问回答即可。稍微注意一点就是 $pre$ 数组初识应该全为 $INF$ 防止影响 $<l$ 的最小 $pre$ 查询。需要注意防止误删，因为我们的删除本质是把上一个 $pre$ 处的 $pre$ 值赋值为 $INF$，如果它前面没有数应该为 $0$。这个时候显然线段树如果你写的 $[1,n]$ 的就没办法将 $0$ 处去设置成 $INF$。所以最好的办法还是不用管没有 $pre$ 的。换句话来讲，保证删除的 $pre$ 在 $1\sim n$ 中即可。

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 = 5e5 + 10;
constexpr int INF = 5e5 + 5;
int a[N];
 
struct Node
{
    int minPre = INF;
    int minVal;
} node[N << 2];
 
#define minPre(x) node[x].minPre
#define minVal(x) node[x].minVal
 
inline void push_up(const int curr)
{
    minPre(curr) = min(minPre(ls(curr)),minPre(rs(curr)));
    minVal(curr) = minPre(curr) == minPre(ls(curr)) ? minVal(ls(curr)) : minVal(rs(curr));
}
 
int n, q;
 
inline void cover(const int curr, const int pos, const int newPos, const int val, const int l = 1, const int r = n)
{
    const int mid = l + r >> 1;
    if (l == r)
    {
        minPre(curr) = newPos, minVal(curr) = val;
        return;
    }
    if (pos <= mid)cover(ls(curr), pos, newPos, val, l, mid);
    else cover(rs(curr), pos, newPos, val, mid + 1, r);
    push_up(curr);
}
 
inline void merge(pii& curr, const pii& other)
{
    if (other.first < curr.first)curr = other;
}
 
inline pii query(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (l <= s and e <= r)return pii(minPre(curr),minVal(curr));
    const int mid = s + e >> 1;
    pii ans(INF, 0);
    if (l <= mid)merge(ans, query(ls(curr), l, r, s, mid));
    if (r > mid)merge(ans, query(rs(curr), l, r, mid + 1, e));
    return ans;
}
 
vector<pii> seg[N];
int ans[N];
int pre[N];
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i];
    cin >> q;
    forn(i, 1, q)
    {
        int l, r;
        cin >> l >> r;
        seg[r].emplace_back(i, l);
    }
    forn(r, 1, n)
    {
        if (pre[a[r]])cover(1, pre[a[r]], INF, INF);//注意保证pre在[1,n]中
        cover(1, r, pre[a[r]], a[r]);
        pre[a[r]] = r;
        for (const auto [id,l] : seg[r])
        {
            auto [MinPre, val] = query(1, l, r);
            if (MinPre < l)ans[id] = val;
        }
    }
    forn(i, 1, q)cout << ans[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;
}
 

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O((n+q)\log n)$$