Submission #1068343

#	Time	Username	Problem	Language	Result	Execution time	Memory
1068343		ewirlan	Overtaking (IOI23_overtaking)	C++17	39 / 100	3569 ms	46164 KiB

overtaking

//
#ifndef __SIZEOF_INT128__
  #define __SIZEOF_INT128__
#endif
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
template <typename T> using oset =  tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define rep(i, p, k) for(int i(p); i < (k); ++i)
#define per(i, p, k) for(int i(p); i > (k); --i)
#define sz(x) (int)(x).size()
#define sc static_cast
typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef __int128_t lll;
//#define int ll
template <typename T = int> using par = std::pair <T, T>;
#define fi first
#define se second
#define test int _number_of_tests(in()); while(_number_of_tests--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb emplace_back
struct Timer {
    string name{""};
    time_point<high_resolution_clock> end, start{high_resolution_clock::now()};
    duration<float, std::milli> dur;
    Timer() = default;
    Timer(string nm): name(nm) {}
    ~Timer() {
        end = high_resolution_clock::now(); dur= end - start;
        cout << "@" << name << "> " << dur.count() << " ms" << '\n';
    }
};
template <typename T = int> inline T in()
{
    static T x;
    std::cin >> x;
    return x;
}
std::string yn(bool b)
{
    if(b) return "YES\n";
    else return "NO\n";
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par);
template <typename T> std::ostream& operator<< (std::ostream& out, const std::vector <T>& wek)
{
    for(const auto& i : wek)out << i << ' ';
    return out;
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par)
{
    out << '{'<<par.first<<", "<<par.second<<"}";
    return out;
}
#define show(x) cerr << #x << " = " << x << '\n';
int l, n, x, m;
vector <ll> t;
vector <int> w;
vector <int> s;
#define g0 get<0>
#define g1 get<1>
#define g2 get<2>
constexpr int maxn = 1003;
vector <tuple <ll, int, int>> dp[maxn];
ll mt[maxn][maxn];
map <ll, ll> odp;
struct ty{
    ll p, k, w;
    ty(ll a, ll b, ll c) : p(a), k(b), w(c) {}
};
vector <ty> tst[maxn];
void init(int Wl, int Wn, vector <ll> Wt, vector <int> Ww, int Wx, int Wm, vector <int> Ws)
{
    l = Wl;
    n = Wn;
    t = Wt;
    x = Wx;
    m = Wm;
    w = Ww;
    s = Ws;
    vector < tuple<ll, int, int>> v;
    rep(i, 0, n)v.pb(t[i], w[i], i);
    rep(i, 1, m){
        sort(all(v));
        rep(j, 0, n)mt[i-1][g2(v[j])] = g0(v[j]);
        dp[i-1] = v;
        ll d(s[i]-s[i-1]);
        vector <tuple <ll, int, int>> v2;
        v2.pb(g0(v[0]) + g1(v[0])*d, g1(v[0]), g2(v[0]));
        rep(j, 1, n)v2.pb(max(g0(v[j])+ g1(v[j])*d, g0(v2[j-1])), g1(v[j]), g2(v[j]));
        v = v2;
    }
    rep(j, 0, n)mt[m-1][g2(v[j])] = g0(v[j]);
    dp[m-1] = v;
    // rep(i, 0, m+1)dp[i].pb(1e18, 0, n+1);
    // rep(i, 0, m+1)mt[i][n+1] = 1e18;
    rep(i, 1, m){
        ll d(x*ll(s[i]-s[i-1]));
        vector <pair <ll, ll>> v;
        rep(j, 0, n)v.pb(g0(dp[i-1][j]) + (g1(dp[i-1][j])<=x?0:1), mt[i][g2(dp[i-1][j])]-d);
        vector <ty> w;
        for(auto [g, m]: v){
            if(sz(w) && w.back().w >= m)continue;
            while(sz(w) && (w.back().p == g))w.pop_back();
            if(sz(w)) w.back().k = min(w.back().k, g-1);
            w.pb(g, m, m);
        }
        tst[i] = w;
        // cerr << "--------------------------\n";
        // for(auto [p, k, w]: w)cerr << p << '-' << k << ": " << w << '\n';
    }
}
ll arrival_time(ll y)
{
    rep(i, 1, m){
        ll d(x*ll(s[i]-s[i-1]));
        vector <ll> dm;
        for(auto [p, k, w]: tst[i])if(p <= y && y <= k)dm.pb(w);
        for(auto i: dm)y = max(y, i);
        y += d;
    }
    return y;
}

• Subtask 0: 0 / 0
• Subtask 1: 9 / 9
• Subtask 2: 10 / 10
• Subtask 3: 20 / 20
• Subtask 4: 0 / 26
• Subtask 5: 0 / 35