제출 #96838

#제출 시각아이디문제언어결과실행 시간메모리
96838KastandaICC (CEOI16_icc)C++11
61 / 100
218 ms760 KiB
// I do it for the glory
#include<bits/stdc++.h>
#include "icc.h"
#define pb push_back
using namespace std;
mt19937 Rnd(time(0));
int fe, se;
bool Rand()
{
    return (Rnd() & 1);
}
void GoForIt(vector < int > L, vector < int > R)
{
    while (L.size() > 1)
    {
        vector < int > L1, L2;
        shuffle(L.begin(), L.end(), Rnd);
        for (int i = 0; i < (int)L.size(); i++)
        {
            if (i < ((int)L.size() >> 1))
                L1.pb(L[i]);
            else
                L2.pb(L[i]);
        }
        assert(abs((int)L1.size() - (int)L2.size()) <= 1);
        bool w = query(L1.size(), R.size(), L1.data(), R.data());
        if (w) L = L1;
        else L = L2;
    }
    while (R.size() > 1)
    {
        vector < int > R1, R2;
        shuffle(R.begin(), R.end(), Rnd);
        for (int i = 0; i < (int)R.size(); i++)
        {
            if (i < ((int)R.size() >> 1))
                R1.pb(R[i]);
            else
                R2.pb(R[i]);
        }
        assert(abs((int)R1.size() - (int)R2.size()) <= 1);
        bool w = query(L.size(), R1.size(), L.data(), R1.data());
        if (w) R = R1;
        else R = R2;
    }
    fe = L[0];
    se = R[0];
    return ;
}
void run(int N)
{
    int P[N + 1];
    vector < int > A[N + 1];
    for (int i = 1; i <= N; i++)
        A[i].pb(i), P[i] = i;
    for (int q = 1; q <= N - 1; q ++)
    {
        bool done = 0;
        vector < int > L, R;
        while (!done)
        {
            L.clear(); R.clear();
            for (int i = 1; i <= N; i++)
            {
                bool w = Rand();
                for (int v : A[i])
                {
                    if (w) L.pb(v);
                    else R.pb(v);
                }
            }
            if (!L.size() || !R.size())
                continue; // What are the odds of that ?
            done = query(L.size(), R.size(), L.data(), R.data());
        }
        GoForIt(L, R);
        setRoad(fe, se);
        int v = fe, u = se, pu = P[u];
        for (int w : A[pu])
            P[w] = P[v], A[P[v]].pb(w);
        A[pu].clear();
    }
    return ;
}
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...