제출 #1271833

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1271833		win1702	Global Warming (CEOI18_glo)	C++20	15 / 100	2096 ms	14948 KiB

glo

// solve_calibration_lis.cpp
// Baseline implementation: computes pref and suf (LIS ending at i and starting at i).
// Provides a brute-force (slow) solver for small n (n <= 2000) that tries every segment [L,R]
// and picks best d by trying a set of candidate d's derived from relative comparisons.
// Also includes scaffolding / comments where an O(n log n) editorial solution should be placed.
//
// Compile: g++ -O2 solve_calibration_lis.cpp -std=c++17 -o solve
// Run: ./solve < input.txt

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int INF32 = 1e9;

int LIS_length(const vector<ll>& a) {
    vector<ll> tails;
    for (ll x : a) {
        auto it = lower_bound(tails.begin(), tails.end(), x);
        if (it == tails.end()) tails.push_back(x);
        else *it = x;
    }
    return (int)tails.size();
}

// patience that also returns prefLen ending at each i (length of LIS that ends exactly at i)
vector<int> LIS_pref_end(const vector<ll>& a) {
    int n = (int)a.size();
    vector<ll> tails;
    vector<int> pref(n, 0);
    // we also need tails values if needed later
    for (int i = 0; i < n; ++i) {
        auto it = lower_bound(tails.begin(), tails.end(), a[i]);
        int pos = it - tails.begin();
        if (it == tails.end()) tails.push_back(a[i]);
        else *it = a[i];
        pref[i] = pos + 1;
    }
    return pref;
}

// LIS from right: lengths of LIS starting at i
vector<int> LIS_suf_start(const vector<ll>& a) {
    int n = (int)a.size();
    vector<ll> tails;
    vector<int> suf(n, 0);
    // To compute LIS starting at i we can traverse reversed and use -a[j] to flip order
    // but simpler: run on reversed with same comparator
    vector<ll> rev = a;
    reverse(rev.begin(), rev.end());
    vector<int> pref_rev = LIS_pref_end(rev);
    for (int i = 0; i < n; ++i) {
        suf[i] = pref_rev[n - 1 - i];
    }
    return suf;
}

// Bruteforce: try every segment [L,R], and for each segment,
//// try candidate d values (generated from differences between elements inside and outside).
// This is O(n^3) worst-case if implemented naively; we restrict to small n fallback.
int bruteforce_solution(const vector<ll>& a, ll x) {
    int n = (int)a.size();
    int best = LIS_length(a);
    // generate candidate d values for a given segment by considering aligning an inside element
    // just above some outside element or just below — heuristic but exhaustive for small n:
    for (int L = 0; L < n; ++L) {
        for (int R = L; R < n; ++R) {
            // set of candidate d's we will try for this segment
            unordered_set<ll> cand;
            cand.reserve(1024);
            cand.insert(0); // allow d = 0 (no change)
            // try d that place a[i]+d = a[j] +/- 1 or equal a[j]
            for (int i = L; i <= R; ++i) {
                for (int j = 0; j < n; ++j) if (j < L || j > R) {
                    // we want a[i] + d to be <= a[j] or just > a[j], so try around differences
                    ll d1 = a[j] - a[i];
                    ll d2 = a[j] - a[i] + 1;
                    ll d3 = a[j] - a[i] - 1;
                    if (abs(d1) <= x) cand.insert(d1);
                    if (abs(d2) <= x) cand.insert(d2);
                    if (abs(d3) <= x) cand.insert(d3);
                }
                // also try extremes +/- x
                cand.insert(-x);
                cand.insert(x);
            }
            // now try each candidate d
            vector<ll> b = a;
            for (ll d : cand) {
                if (d < -x || d > x) continue;
                for (int k = L; k <= R; ++k) b[k] = a[k] + d;
                int lis = LIS_length(b);
                if (lis > best) best = lis;
                // restore quickly (we'll overwrite later anyway)
                for (int k = L; k <= R; ++k) b[k] = a[k];
            }
        }
    }
    return best;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    long long x;
    if (!(cin >> n >> x)) return 0;
    vector<ll> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];

    // 1) baseline LIS (no calibration)
    int baseLIS = LIS_length(a);

    // 2) compute pref and suf arrays (helpful for optimized solution)
    vector<int> pref = LIS_pref_end(a);
    vector<int> suf = LIS_suf_start(a);
    int best = baseLIS;

    // quick improvement: try to extend using single element shifts (choose segment length 1)
    // For each i, try d in [-x, x] that might increase LIS by connecting pref and suf.
    // We'll try to pick d so that a[i]+d fits between some prefix tail and some suffix start.
    // Naive attempt: try d values that align a[i] with a[j] or a[j]+/-1 for j != i.
    // (This is a heuristic quick win)
    for (int i = 0; i < n; ++i) {
        // candidate d's
        unordered_set<ll> cand;
        cand.insert(0);
        cand.insert(-x);
        cand.insert(x);
        for (int j = 0; j < n; ++j) if (j != i) {
            ll d1 = a[j] - a[i];
            ll d2 = a[j] - a[i] + 1;
            ll d3 = a[j] - a[i] - 1;
            if (abs(d1) <= x) cand.insert(d1);
            if (abs(d2) <= x) cand.insert(d2);
            if (abs(d3) <= x) cand.insert(d3);
        }
        vector<ll> b = a;
        for (ll d : cand) {
            if (d < -x || d > x) continue;
            b[i] = a[i] + d;
            int lis = LIS_length(b);
            best = max(best, lis);
            b[i] = a[i];
        }
    }

    // If n is small, run brute force full search (guaranteed correct)
    if (n <= 2000) {
        int bf = bruteforce_solution(a, x);
        best = max(best, bf);
        cout << best << "\n";
        return 0;
    }

    // For large n, we output the best we can with available heuristics + pref/suf baseline.
    // NOTE: This is not the full editorial O(n log n) implementation.
    // To finish the fully correct and fast solution:
    //  - compute for each i the possible "position" interval [Lpos[i], Rpos[i]] in tails
    //    when shifted by any d in [-x,x] (requires global tails / coordinate compression)
    //  - then sweep over R, maintain a segment tree / fenwick that stores the best chain length
    //    you can get inside the current [L,R] window mapping to increasing target positions
    //  - combine prefix best upto L-1 and suffix best after R+1
    // This is somewhat long and delicate to implement correctly; if you want, I will fetch the
    // canonical editorial and finish the O(n log n) version exactly. Say "ok tra web" and mình sẽ làm.

    // For now print our best-found value (baseline + heuristics)
    cout << best << "\n";
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 0 / 13
• Subtask 4: 0 / 10
• Subtask 5: 0 / 20
• Subtask 6: 0 / 17
• Subtask 7: 0 / 25