Submission #1212359

#	Time	Username	Problem	Language	Result	Execution time	Memory
1212359		countless	Nile (IOI24_nile)	C++20	36 / 100	2095 ms	10688 KiB

nile

#include "nile.h"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

const ll MOD = 998244353;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp << " " <<
#define REP(i, a, b) for (ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

vector<ll> calculate_costs(vector<int> w, vector<int> a, vector<int> b, vector<int> e) {
    // fixing cause my code is so messy
    
    // pre
    int n = w.size(), q = e.size(), targ;
    ll diff, prev = 0, curr;

    // dec
    vector<bool> vis(n, false);
    vector<int> p(n), s(n, 1), m(n, 1e9), c(n), o(n), eq(q);
    vector<pair<int, int>> mm(n, {1e9, -1e9});
    vector<ll> r(q);
    vector<pair<int, pair<int, int>>> st;
    set<int> rele;
    
    // helpe
    auto eval = [&](int i, int j) -> int {
        return abs(w[o[i]] - w[o[j]]);
    };

    auto find = [&](auto &&find, int u) -> int {
        if (u == p[u]) return u;
        return p[u] = find(find, p[u]);
    };

    auto unite = [&](int u, int v) -> void {
        int x = u, y = v;
        u = find(find, u), v = find(find, v);

        mm[u].fi = min(mm[u].fi, x);
        mm[u].se = max(mm[u].se, y);

        int parity;
        parity = (x - mm[u].fi) % 2;
        if (!parity or (parity and (eval(x-1, x+1) <= targ))) m[u] = min(m[u], c[o[x]]);

        parity = (y - mm[u].fi) % 2;
        if (!parity or (parity and (eval(y-1, y+1) <= targ))) m[u] = min(m[u], c[o[y]]);

        m[u] = min(m[u], m[v]);

        if (u == v) return;

        if (!vis[o[x]]) diff -= c[o[x]];
        if (!vis[o[y]]) diff -= c[o[y]];


        vis[o[x]] = true, vis[o[y]] = true;

        p[v] = u;
        s[u] += s[v];

        if (s[u] % 2 == 1) rele.insert(u);
        else { auto it = rele.find(u); if (it != rele.end()) rele.erase(it); }
        auto it = rele.find(v);
        if (it != rele.end()) rele.erase(it);
        return;
    };

    // proc
    REP(i, 0, n) c[i] = a[i] - b[i];
    iota(all(o), 0), iota(all(eq), 0), iota(all(p), 0);
    sort(all(o), [&](const int i, const int j) -> bool {
        return w[i] < w[j];
    });

    sort(all(eq), [&](const int i, const int j) -> bool {
        return e[i] < e[j];
    });

    REP(i, 0, n) {
        if (i < n-1) st.push_back({eval(i, i+1), {i, i+1}});
        if (i < n-2) st.push_back({eval(i, i+2), {i, i+2}});
    }

    sort(rall(st));

    // comp
    REP(qq, 0, q) {
        targ = e[eq[qq]];
        diff = curr = 0;

        // it's monotonic for a given e[i], so compute per new e[i] ! ! !
        // treat like stack

        // compute previous state then subtract diff
        while (!st.empty() and st.back().first <= targ) {
            auto [ww, edg] = st.back();
            unite(edg.fi, edg.se);
            st.pop_back();
        }

        for (auto &cc : rele) {
            curr += m[cc];
            // cerr << cc sp m[cc] << endl;
        }

        r[eq[qq]] = (qq > 0 ? r[eq[qq-1]] : accumulate(all(a), 0LL)) + diff - prev + curr;
        prev = curr;
    }

    return r;
}

• Subtask 0: 0 / 0
• Subtask 1: 6 / 6
• Subtask 2: 13 / 13
• Subtask 3: 17 / 17
• Subtask 4: 0 / 11
• Subtask 5: 0 / 20
• Subtask 6: 0 / 15
• Subtask 7: 0 / 18