제출 #1203964

#제출 시각아이디문제언어결과실행 시간메모리
1203964Zbyszek99가장 긴 여행 (IOI23_longesttrip)C++20
100 / 100
7 ms432 KiB
#include "longesttrip.h"
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define ld long double
#define ull unsigned long long
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define rep(i, b) for(int i = 0; i < (b); ++i)
#define rep2(i,a,b) for(int i = a; i <= (b); ++i)
#define rep3(i,a,b,c) for(int i = a; i <= (b); i+=c)
#define count_bits(x) __builtin_popcountll((x))
#define all(x) (x).begin(),(x).end()
#define siz(x) (int)(x).size()
#define forall(it,x) for(auto& it:(x))
using namespace __gnu_pbds;
using namespace std;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
mt19937 mt;void random_start(){mt.seed(chrono::time_point_cast<chrono::milliseconds>(chrono::high_resolution_clock::now()).time_since_epoch().count());}
ll los(ll a, ll b) {return a + (mt() % (b-a+1));}
const int INF = 1e9+50;
const ll INF_L = 1e18+40;
const ll MOD = 1e9+7;

vi longest_trip(int n, int D)
{
    random_start();
    vi verts;
    rep(i,n) verts.pb(i);
    shuffle(all(verts),mt);
    vi t1 = {verts[0]};
    vi t2 = {verts[1]};
    bool is_nc = 0;
    rep2(i,2,n-1)
    {
        // forall(it,t1) cout << it << " ";
        // cout << "t1\n";
        // forall(it,t2) cout << it << " ";
        // cout << "t2\n\n";
        int v = verts[i];
        if(los(0,1) == 0) swap(t1,t2);
        if(are_connected({v},{t1.back()}))
        {
            t1.pb(v);
            is_nc = 0;
        }
        else
        {
            if(is_nc)
            {
                is_nc = 1;
                t2.pb(v);
            }
            else
            {
                if(are_connected({v},{t2.back()}))
                {
                    is_nc = 1;
                    t2.pb(v);
                }
                else
                {
                    for(int j = siz(t2)-1; j >= 0; j--)
                    {
                        t1.pb(t2[j]);
                    }
                    t2 = {v};
                    is_nc = 0;
                }
            }
        }
    }
    vi ends1;
    vi ends2;
    if(siz(t1) == 1) ends1 = {t1[0]};
    else ends1 = {t1[0],t1.back()};
    if(siz(t2) == 1) ends2 = {t2[0]};
    else ends2 = {t2[0],t2.back()};
    if(!are_connected(t1,t2))
    {
        if(siz(t1) > siz(t2)) return t1;
        return t2;
    }
    if(are_connected(ends1,ends2))
    {
        if(are_connected(ends1,{ends2[0]}))
        {
            if(are_connected({ends1[0]},{ends2[0]}))
            {
                reverse(all(t1));
                forall(it,t2) t1.pb(it);
                return t1;
            }
            else
            {
                forall(it,t2) t1.pb(it);
                return t1;
            }
        }
        else
        {
            reverse(all(t2));
            if(are_connected({ends1[0]},{ends2.back()}))
            {
                reverse(all(t1));
                forall(it,t2) t1.pb(it);
                return t1;
            }
            else
            {
                forall(it,t2) t1.pb(it);
                return t1;
            }
        }
    }
    int l = 0;
    int r = siz(t1)-2;
    int ans = t1[siz(t1)-1];
    while(l <= r)
    {
        int mid = (l+r)/2;
        vi new_t1;
        rep(i,mid+1) new_t1.pb(t1[i]);
        if(are_connected(new_t1,t2))
        {
            ans = t1[mid];
            r = mid-1;
        }
        else
        {
            l = mid+1;
        }
    }
    l = 0;
    r = siz(t2)-2;
    int ans2 = t2[siz(t2)-1];
    while(l <= r)
    {
        int mid = (l+r)/2;
        vi new_t2;
        rep(i,mid+1) new_t2.pb(t2[i]);
        if(are_connected({ans},new_t2))
        {
            ans2 = t2[mid];
            r = mid-1;
        }
        else
        {
            l = mid+1;
        }
    }
    vi nt1;
    int p1 = 0;
    rep(i,siz(t1))
    {
        if(t1[i] == ans) p1 = i;
    }
    rep(i,siz(t1))
    {
        nt1.pb(t1[p1]);
        p1++;
        p1 %= siz(t1);
    }
    reverse(all(nt1));
    int p2 = 0;
    rep(i,siz(t2))
    {
        if(t2[i] == ans2) p2 = i;
    }
    rep(i,siz(t2))
    {
        nt1.pb(t2[p2]);
        p2++;
        p2 %= siz(t2);
    }
    return nt1;
}
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