제출 #1170242

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1170242		Nelt	Palembang Bridges (APIO15_bridge)	C++20	100 / 100	152 ms	12892 KiB

bridge

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define endl "\n"
using namespace std;
using namespace __gnu_pbds;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;

void solve()
{
    ll n, k;
    cin >> k >> n;
    vector<pair<ll, ll>> a;
    ll res = 0, sum = 0;
    for (ll i = 0; i < n; i++)
    {
        ll x, y;
        char c, d;
        cin >> c >> x >> d >> y;
        sum += abs(x - y);
        if (c != d)
            a.push_back(make_pair(min(x, y), max(x, y)));
        else
            res += abs(x - y);
    }
    n = a.size();
    if (n == 0)
    {
        cout << res << endl;
        return;
    }
    if (k == 1)
    {
        vector<ll> l, r;
        for (auto i : a)
        l.push_back(i.first), r.push_back(i.second);
        sort(l.begin(), l.end());
        sort(r.begin(), r.end());
        ll ans = 2e18;
        ll pref[n], pref1[n];
        for (ll i = 0; i < n; i++)
            pref[i] = (i == 0 ? 0 : pref[i - 1]) + l[i];
        for (ll i = 0; i < n; i++)
            pref1[i] = (i == 0 ? 0 : pref1[i - 1]) + r[i];
        for (auto _i : a)
        {
            for (ll x : {_i.first, _i.second})
            {
                ll pos = upper_bound(l.begin(), l.end(), x) - l.begin(), pos1 = lower_bound(r.begin(), r.end(), x) - r.begin();
                ans = min(ans, ((pref[n - 1] - (pos - 1 < 0 ? 0 : pref[pos - 1])) - x * (n - pos) + x * pos1 - (pos1 - 1 < 0 ? 0 : pref1[pos1 - 1])) * 2 + sum + n);
            }
        }
        cout << ans << endl;
        return;
    }
    sort(a.begin(), a.end(), [&](pair<ll, ll> a, pair<ll, ll> b)
         { return a.first + a.second < b.first + b.second; });
    ll pref[n], suf[n];
    multiset<ll> l, r;
    ll suml = 0, sumr = 0;
    for (ll i = 0; i < n; i++)
    {
        l.insert(a[i].first);
        l.insert(a[i].second);
        suml += a[i].first + a[i].second;
        while (!l.empty() and !r.empty() and *l.rbegin() > *r.begin())
        {
            suml -= *l.rbegin();
            sumr += *l.rbegin();
            r.insert(*l.rbegin());
            l.erase(--l.end());
        }
        while (l.size() > r.size()) 
        {
            suml -= *l.rbegin();
            sumr += *l.rbegin();
            r.insert(*l.rbegin());
            l.erase(--l.end());
        }
        while (r.size() > l.size())
        {
            suml += *r.begin();
            sumr -= *r.begin();
            l.insert(*r.begin());
            r.erase(r.begin());
        }
        pref[i] = sumr - suml;
    }
    l.clear();
    r.clear();
    suml = sumr = 0;
    for (ll i = n - 1; i >= 0; i--)
    {
        l.insert(a[i].first);
        l.insert(a[i].second);
        suml += a[i].first + a[i].second;
        while (!l.empty() and !r.empty() and *l.rbegin() > *r.begin())
        {
            suml -= *l.rbegin();
            sumr += *l.rbegin();
            r.insert(*l.rbegin());
            l.erase(--l.end());
        }
        while (l.size() > r.size()) 
        {
            suml -= *l.rbegin();
            sumr += *l.rbegin();
            r.insert(*l.rbegin());
            l.erase(--l.end());
        }
        while (r.size() > l.size())
        {
            suml += *r.begin();
            sumr -= *r.begin();
            l.insert(*r.begin());
            r.erase(r.begin());
        }
        suf[i] = sumr - suml;
    }
    ll ans = pref[n - 1] + res + n;
    for (ll i = 0; i + 1 < n; i++) ans = min(ans, pref[i] + suf[i + 1] + res + n);
    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll t = 1;
    // precomp();
    // cin >> t;
    for (ll cs = 1; cs <= t; cs++)
        solve();
    // cerr << "\nTime elapsed: " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n";
}

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