Submission #1260563

#	Time	Username	Problem	Language	Result	Execution time	Memory
1260563		ewirlan	Arranging Tickets (JOI17_arranging_tickets)	C++20	85 / 100	157 ms	13672 KiB

arranging_tickets

//
#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';
struct prt{ int a, b, c; };
vector <prt> v;
int t;
constexpr int maxn = 2e5 + 3;
vector <int> zp[maxn];
vector <ll> pr;
struct kult
{
    int b, c, i;
};
bool operator<(kult a, kult b){return a.b < b.b;}
int n;
bool da(int x)
{
    // cerr << "da(" << x << ")\n";
    // rep(i, 1, n+1)cerr << pr[i] << ' ';
    // cerr << '\n';
    for(auto odw: {pr[t]-x, pr[t]-x+1}){
        vector <pair <int, int>> zm;
        if(odw < 0)continue;
        ll us(0);
        priority_queue <kult> kul;
        vector <ll> pr2(n+1);
        bool d(1);
        rep(i, 1, t+1){
            for(auto j: zp[i])kul.push(kult{v[j].b, v[j].c, j});
            int z((pr[i] + odw - x - 2*us + 1)/2);
            if(i == t)z = odw - us;
            // cerr << i << ": " << z << '\n';
            while(z > 0){
                if(!sz(kul))goto nie;
                auto [b, c, j] = kul.top(); kul.pop();
                if(c > z){
                    us += z;
                    zm.pb(j, z);
                    kul.push(kult{b, c-z, j});
                    z = 0;
                }
                else{
                    us += c;
                    zm.pb(j, c);
                    z -= c;
                }
            }
            // cerr << "zm: " << zm << '\n';
        }
        for(auto [i,c]: zm){
            pr2[1] += c;
            pr2[v[i].a] -= 2*c;
            pr2[v[i].b+1] += 2*c;
        }
        rep(i, 1, n+1)pr2[i] += pr2[i-1];
        rep(i, 1, n+1)d &= pr2[i] + pr[i] <= x;
        if(d)return 1;
        nie:;
    }
    return 0;
}
std::int32_t main()
{
    std::cout.tie(nullptr); //for luck
    std::cin.tie(nullptr); std::ios_base::sync_with_stdio(0);
    cin >> n;
    int m(in());
    rep(i, 0, m){
        int a(in()), b(in());
        v.pb(prt{min(a, b), max(a,b)-1, in()});
    }
    pr.resize(n+1);
    rep(i, 0, m){
        pr[v[i].a] += v[i].c;
        pr[v[i].b+1] -= v[i].c;
    }
    rep(i, 1, n+1)pr[i] += pr[i-1];
    rep(i, 1, n+1)if(pr[i] >= pr[t])t = i;
    rep(i, 0, m)if(v[i].a <= t && t <= v[i].b)zp[v[i].a].pb(i);
    int p(0), k(pr[t]);
    while(k > p){
        int s((p+k)/2);
        if(da(s))k = s;
        else p = s+1;
    }
    cout << p << '\n';
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 35 / 35
• Subtask 3: 20 / 20
• Subtask 4: 20 / 20
• Subtask 5: 0 / 15