제출 #39508

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
39508		qoo2p5	Palembang Bridges (APIO15_bridge)	C++14	22 / 100	86 ms	11988 KiB

bridge

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int INF = (int) 1e9 + 123;
const ll LINF = (ll) 1e18 + 123;

#define rep(i, s, t) for (auto i = (s); i < (t); ++(i))
#define per(i, s, t) for (auto i = (s); i >= (t); --(i))
#define sz(x) ((int) (x).size())
#define mp make_pair
#define pb push_back
#define all(x) (x).begin(), (x).end()

bool mini(auto &x, const auto &y) {
	if (y < x) {
		x = y;
		return 1;
	}
	return 0;
}

bool maxi(auto &x, const auto &y) {
	if (y < x) {
		x = y;
		return 1;
	}
	return 0;
}

void run();

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	run();
	return 0;
}

const int N = (int) 1e5 + 123;

int k, n;
vector<pair<ll, ll>> a;
vector<ll> cool;
ll total;

void solve1() {
	vector<ll> p;
	{
		for (auto &it : a) {
			p.pb(it.first);
			p.pb(it.second);
		}
		sort(all(p));
	}
	
	int k = sz(p) / 2;
	ll x = p[k];
	ll ans = 0;
	for (ll y : p) ans += abs(x - y);
	
	ans += total;
	cout << ans << "\n";
}

ll pref[N], suff[N];

ll solve() {
	sort(all(a), [] (const pair<ll, ll> &x, const pair<ll, ll> &y) {
		return x.first + x.second < y.first + y.second;
	});
	
	{
		ll sum_l = 0, sum_r = 0;
		multiset<ll> L, R;
		multiset<ll> q;
		rep(i, 0, n) {
			R.insert(a[i].first);
			R.insert(a[i].second);
			sum_r += a[i].first + a[i].second;
			
			q.insert(a[i].first);
			q.insert(a[i].second);
			while (sz(q) > (sz(L) + sz(R) + 1) / 2) {
				q.erase(q.find(*q.rbegin()));
			}
			ll mid = *q.rbegin();
			
			while (sz(R) && *R.begin() <= mid) {
				ll el = *R.begin();
				sum_r -= el;
				sum_l += el;
				L.insert(el);
				R.erase(R.find(el));
			}
			
			pref[i] = (mid * sz(L) - sum_l) + (sum_r - mid * sz(R));
		}
	}
	
	{
		ll sum_l = 0, sum_r = 0;
		multiset<ll> L, R;
		multiset<ll> q;
		per(i, n - 1, 0) {
			R.insert(a[i].first);
			R.insert(a[i].second);
			sum_r += a[i].first + a[i].second;
			
			q.insert(a[i].first);
			q.insert(a[i].second);
			while (sz(q) > (sz(L) + sz(R) + 1) / 2) {
				q.erase(q.find(*q.rbegin()));
			}
			ll mid = *q.rbegin();
			
			while (sz(R) && *R.begin() <= mid) {
				ll el = *R.begin();
				sum_r -= el;
				sum_l += el;
				L.insert(el);
				R.erase(R.find(el));
			}
			
			suff[i] = (mid * sz(L) - sum_l) + (sum_r - mid * sz(R));
		}
	}
	
	ll ans = LINF;
	rep(i, 0, n) {
		mini(ans, pref[i] + suff[i + 1]);
	}
	return ans;
}

void run() {
	cin >> k >> n;
	rep(i, 0, n) {
		char t1, t2;
		ll x1, x2;
		cin >> t1 >> x1 >> t2 >> x2;
		if (t1 == t2) {
			total += abs(x1 - x2);
		} else {
			if (x1 > x2) {
				swap(t1, t2);
				swap(x1, x2);
			}
			a.pb({x1, x2});
			cool.pb(x1);
			cool.pb(x2);
		}
	}
	sort(all(cool));
	cool.resize(unique(all(cool)) - cool.begin());
	sort(all(a));
	n = sz(a);
	total += n;
	if (k == 1) {
		solve1();
		return;
	}
	if (n == 0) {
		cout << total << "\n";
		return;
	}
	
	ll ans = solve();
	
	ans += total;
	cout << ans << "\n";
}

• Subtask 1: 8 / 8
• Subtask 2: 14 / 14
• Subtask 3: 0 / 9
• Subtask 4: 0 / 32
• Subtask 5: 0 / 37