제출 #1171414

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1171414		Bzslayed	Circle Passing (EGOI24_circlepassing)	C++20	100 / 100	75 ms	8688 KiB

Main

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace std;
using namespace __gnu_pbds; 

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

#define ll long long
#define ull unsigned long long
#define llll __int128
#define ld long double
#define pll pair<ll, ll>
#define pii pair<int, int>
#define coutpair(p) cout << p.first << " " << p.second << "|"

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

ll n, m, q;
ll dist(ll x, ll y){ //minimum distance
    return min(abs(x-y), 2*n-abs(x-y));
}

ll frd(ll x){
    if (x >= n) x -= n;
    else x += n;
    return x;
}

ll mndist(ll x, ll y, ll pos1, ll pos2){
    return min(dist(x, pos1)+dist(y, frd(pos1)), dist(x, pos2)+dist(y, frd(pos2)))+1;
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m >> q;
    vector<ll> k;
    for (int i=0; i<m; i++){
        ll c; cin >> c;
        k.push_back(c);
        k.push_back(c+n);
    }
    sort(k.begin(), k.end());

    for (int i=0; i<q; i++){
        ll x, y; cin >> x >> y;

        ll ans = 0;
        ll pos = upper_bound(k.begin(), k.end(), x)-k.begin(); //closest person with a best friend

        ll ans1 = dist(x, y); //only passing to those adjacent for the entire process
        ll ans2 = 0;
        if (pos == 0) ans2 = mndist(x, y, k[pos], k.back());
        else if (pos == 2*m){
            pos--;
            ans2 = mndist(x, y, k[pos], k.front());
        }
        else ans2 = mndist(x, y, k[pos], k[pos-1]);

        //cout << "pos " << pos << " ";
        ans = min(ans1, ans2);
        //cout << "ans1 " << ans1 << " first " << fi << " second " << se << "\n";

        cout << ans << "\n";
    }
    
    return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 20 / 20
• Subtask 3: 22 / 22
• Subtask 4: 17 / 17
• Subtask 5: 27 / 27