제출 #1069610

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1069610		ewirlan	수천개의 섬 (IOI22_islands)	C++17	57.10 / 100	75 ms	21684 KiB

islands

//
#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';
constexpr int maxn = 1e5 + 3;
vector <pair <int, int>> graf[maxn];
vector <pair <int, int>> rgraf[maxn];
bool us[maxn];
int sto[maxn];
void del(int w){
    if(us[w])return;
    us[w] = 1;
    for(auto [i, j]: rgraf[w])if(!--sto[i])del(i);
}
pair <int, int> ns[maxn];
bool byla[maxn], bylb[maxn], wcyk[maxn];
variant <bool, vector <int>> find_journey(int n, int m, vector <int> u, vector <int> v){
    rep(i, 0, m){
        graf[u[i]].pb(v[i], i);
        rgraf[v[i]].pb(u[i], i);
    }
    rep(i, 0, n)sto[i] = sz(graf[i]);
    int s(0);
    vector <int> ds;
    rep(i, 0, n)if(!sto[i])del(i);
    while(!us[s] && sto[s] == 1){
        del(s);
        for(auto [i, j]: graf[s])if(!us[i]){
            ds.pb(j);
            s = i;
            break;
        }
    }
    if(us[s])return false;
    vector <pair <int, int>> hh;
    for(auto i: graf[s])if(!us[i.first])hh.pb(i);
    assert(sz(hh) >= 2);
    auto a(hh[0]), b(hh[1]);
    rep(w, 0, n)if(!us[w] && w != s)for(auto [i, j]: graf[w])if(!us[i] && w != i){
        ns[w] = {i,j};
        break;
    }
    vector <int> cykl, edc;
    vector <pair <int, int>> dc{a}, db{b};
    byla[s] = bylb[s] = 1;
    while(!byla[dc.back().first]){
        // cerr << dc.back() << ' ';
        byla[dc.back().first] = 1;
        dc.pb(ns[dc.back().first]);
    }
    // cerr << '\n';
    while(dc.size() && !wcyk[dc.back().first]){
        cykl.pb(dc.back().first);
        edc.pb(dc.back().second);
        wcyk[dc.back().first] = 1;
        dc.pop_back();
    }
    // cerr << cykl << " : " << edc << '\n';
    while(!bylb[db.back().first] && !wcyk[db.back().first]){
        bylb[db.back().first] = 1;
        db.pb(ns[db.back().first]);
    }
    vector <int> rt;
    for(auto i: ds)rt.pb(i);
    reverse(all(dc));
    rep(xd, 0, 2){
        reverse(all(dc));
        for(auto i: dc)rt.pb(i.second);
        reverse(all(edc)); 
        for(auto i: edc)rt.pb(i);
        reverse(all(dc)); for(auto i: dc)rt.pb(i.second);
        for(auto i: db)rt.pb(i.second);
        if(wcyk[db.back().first]){
            reverse(all(edc));
            rep(i, 0, sz(cykl))if(cykl[i] == db.back().first){
                rep(j, 0, sz(edc))rt.pb(edc[(i+j) % sz(edc)]);
                reverse(all(db));
                for(auto i: db)rt.pb(i.second);
                reverse(all(db));
                break;
            }
        }
        else{
            per(i, sz(db)-2, -1)if(db[i].first == db.back().first){
                per(j, i, -1)rt.pb(db[j].second);
            }
        }
    }
    reverse(all(ds));
    for(auto i: ds)rt.pb(i);
    return rt;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 7.35 / 21
• Subtask 4: 24 / 24
• Subtask 5: 15.749999999999998 / 45