제출 #884477

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
884477		AnhPham	Palembang Bridges (APIO15_bridge)	C++17	31 / 100	33 ms	3832 KiB

bridge

#include <bits/stdc++.h>

#ifdef OP_DEBUG
    #include <algo/debug.h>
#else
    #define debug(...) 26
#endif

using namespace std;

#define int 	long long // maybe tle?
#define sz(v)   (int)(v).size()
#define all(v)  (v).begin(), (v).end()
#define TcT     template <class T

const   int     MOD = (int)1e9 + 7, INF = (int)2e9 + 9, INF64 = (int)4e18 + 18;

TcT>            bool minimize(T &val, const T &upd) { return upd < val ? val = upd, 1 : 0; }
TcT>            bool maximize(T &val, const T &upd) { return upd > val ? val = upd, 1 : 0; }

TcT, class S>   istream &operator >> (istream &scanf, pair <T, S> &u) { return scanf >> u.first >> u.second;        }
TcT, class S>   ostream &operator << (ostream &print, pair <T, S> &u) { return print << u.first << ' ' << u.second; }

void solve();

int32_t main() {
    cin.tie(nullptr), cout.tie(nullptr) -> sync_with_stdio(0);
    int testcases = 1;

#define TC 0
    if (TC) { cin >> testcases; } for (; testcases--;) { solve(); }
}

/* [Pham Hung Anh - 11I - Tran Hung Dao High School for Gifted Student] */

int k, n;
int same_side;
vector <int> buildings;
vector <pair <int, int>> opposite;

namespace sub12 { // O(n)
    void solve() {
        int ans = same_side;
        if (sz(buildings) == 0)
            return void(cout << ans << '\n');

        int m = sz(buildings);
        int median = buildings[(m + 1) / 2];

        for (int i = 0; i < m; ++i)
            ans += abs(median - buildings[i]);
        
        cout << ans + sz(buildings) / 2 << '\n';
    }
}

namespace sub3 { // O(n^3)
    void solve() {
        int ans = same_side;
        if (sz(buildings) == 0)
            return void(cout << ans << '\n');

        int res = INF64;
        for (int i = 0; i < sz(buildings); ++i)
            for (int j = i + 1; j < sz(buildings); ++j) {
                int bridge1 = buildings[i], bridge2 = buildings[j];
                int sum = 0;
                for (auto p : opposite)
                    sum += min(abs(p.first - bridge1) + abs(p.second - bridge1), abs(p.first - bridge2) + abs(p.second - bridge2)) + 1;
                
                minimize(res, sum);
            }
            
        cout << ans + res << '\n';
    }
}

namespace sub4 { // O(log3(n)^2 * n)
    int get(int bridge1, int bridge2) {
        int sum = same_side;
        for (auto p : opposite)
            sum += min(abs(p.first - bridge1) + abs(p.second - bridge1), abs(p.first - bridge2) + abs(p.second - bridge2)) + 1;
        
        return sum;
    }

    int ternary_search(int bridge1) {
        int res = INF64;
        int lo = bridge1 + 1, hi = sz(buildings) - 1;

        while (lo <= hi) {
            int mid1 = lo + (hi - lo) / 3;
            int mid2 = hi - (hi - lo) / 3;

            int sum1 = get(buildings[bridge1], buildings[mid1]);
            int sum2 = get(buildings[bridge1], buildings[mid2]);
            minimize(res, min(sum1, sum2));

            if (sum1 < sum2)
                hi = mid2 - 1;
            else if (sum1 > sum2)
                lo = mid1 + 1;
            else
                lo = mid1 + 1, hi = mid2 - 1;
        }

        return res;
    }

    void solve() {
        int ans = INF64;
        int lo = 0, hi = sz(buildings) - 2;
        
        while (lo <= hi) {
            int mid1 = lo + (hi - lo) / 3;
            int mid2 = hi - (hi - lo) / 3;
            
            int ans1 = ternary_search(mid1);
            int ans2 = ternary_search(mid2);
            minimize(ans, min(ans1, ans2));

            if (ans1 < ans2)
                hi = mid2 - 1;
            else if (ans1 > ans2)
                lo = mid1 + 1;
            else
                lo = mid1 + 1, hi = mid2 - 1;
        }

        cout << ans << '\n';
    }
}

namespace sub5 {
    int suml = 0, sumr = 0;
    multiset <int> lo, hi;
    
    void balance() {
        while (sz(lo) > sz(hi)) {
            int x = *lo.rbegin();
            hi.emplace(x);
            suml -= x;
            sumr += x;
            lo.erase(lo.find(x));
        }
    
        while (sz(hi) > sz(lo)) {
            int x = *hi.begin();
            lo.emplace(x);
            suml += x;
            sumr -= x;
            hi.erase(hi.find(x));
        }
    }
    
    void add(int x) {
        lo.emplace(x);
        suml += x;
    
        balance();
    }
    
    void solve() {
        sort(all(opposite), [&](pair <int, int> a, pair <int, int> b) -> bool {
            return a.first + a.second < b.first + b.second;
        });

        vector <int> pref(sz(opposite) + 1);

        for (int i = 0; i < sz(opposite); ++i) {
            add(opposite[i].first);
            add(opposite[i].second);

            pref[i + 1] = sumr - suml;
        }

        int ans = pref[sz(opposite)];
        for (int i = sz(opposite) - 1; i >= 0; --i) {
            add(opposite[i].first);
            add(opposite[i].second);

            minimize(ans, pref[i] + sumr - suml);
        }
    
        cout << ans + same_side << '\n';
    }
}

void solve() {
    cin >> k >> n;

    for (int i = 1; i <= n; ++i) {
        char p, q; int s, t; cin >> p >> s >> q >> t;

        if (p == q)
            same_side += abs(s - t);
        else
            buildings.push_back(s), buildings.push_back(t), 
            opposite.emplace_back(min(s, t), max(s, t));
    }

    sort(all(buildings));

    if (k == 1)
        sub12 :: solve();
    else {
        if (n <= 100)
            sub3 :: solve();
        // else if (n <= 1000)
        //     sub4 :: solve();
        else
            sub5 :: solve();
    }
}

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