Submission #1060609

#TimeUsernameProblemLanguageResultExecution timeMemory
1060609ewirlanHighway Tolls (IOI18_highway)C++17
51 / 100
163 ms11844 KiB
    //
    #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>
    #include "highway.h"
    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 = 90003;
    vector <pair<int, int>> graf[maxn];
    int odl[maxn];
    bool xd[maxn];
    void dfs(int w, bool x)
    {
        odl[w] = 1;
        xd[w] = x;
        for(auto [i, j]: graf[w])if(!odl[i])dfs(i, !x);
    }
    void find_pair(int n, vector <int> ka, vector <int> kb, int a, int b)
    {
        int m(sz(ka));
        rep(i, 0, m){
            graf[ka[i]].pb(kb[i], i);
            graf[kb[i]].pb(ka[i], i);
        }
        dfs(0, 0);
        vector <int> xdd[2];
        rep(i, 0, n)xdd[xd[i]].pb(i);
        vector vv = xdd[!sz(xdd[0]) || (sz(xdd[1]) && sz(xdd[1]) < sz(xdd[0]))];
        ll d(ask(vector<int>(m, 0))/a);
        auto pyt = [n,m](vector <bool> v){
            vector <int> w(m);
            rep(i, 0, n)if(v[i])for(auto [i, j]: graf[i])w[j] |= 1;
            return ask(w);
        };
        int p(0), k(sz(vv)), s;
        vector <bool> v(n);
        while(k > p){
            s = (p+k)/2;
            rep(i, 0, n)v[i] = 0;
            rep(i, 0, sz(vv))v[vv[i]] = i >= s;
            if(pyt(v) == d*a)k = s;
            else p = s+1;
        }
        int r(p-1);
        vector <int> so;
        queue <int> kul; 
        bool je(0);
    rp:
        rep(i, 0, n)odl[i] = 0;
        so = {r};
        kul.push(r);
        odl[r] = 1;
        while(kul.size()){
            auto t(kul.front()); kul.pop();
            for(auto [i, j]: graf[t])if(!odl[i]){
                kul.push(i);
                so.pb(i);
                odl[i] = odl[t]+1;
            }
        }
        if(je){
            vector <int> s;
            for(auto i: so)if(odl[i] == d+1)s.pb(i);
            so = s;
        }
        p = 0, k = sz(so);
        while(k > p){
            s = (p+k)/2;
            rep(i, 0, n)v[i] = 0;
            rep(i, 0, sz(so))v[so[i]] = i >= s;
            if(pyt(v) == d*a)k = s;
            else p = s+1;
        }
        if(!je){
            r = so[p-1];
            je = 1;
            goto rp;
        }
        answer(r, so[p-1]);
    }
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