제출 #526791

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
526791		PurpleCrayon	카니발 티켓 (IOI20_tickets)	C++17	100 / 100	999 ms	134912 KiB

tickets

#include "tickets.h"
#include <bits/stdc++.h>
using namespace std;

#define sz(v) int(v.size())
typedef long long ll;

ll cost(int k, vector<vector<int>> x, vector<vector<int>> s) {
    int n = sz(x), m = sz(x[0]);
    vector<vector<int>> a(k);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (s[i][j] != -1) {
                a[s[i][j]].push_back(x[i][j]);
            }
        }
    }
    ll ans = 0;
    for (auto& v : a) {
        sort(v.begin(), v.end());
        int m = v[sz(v)/2];
        for (auto& c : v) ans += abs(c - m); 
    }
    return ans;
}
ll find_maximum(int k, vector<vector<int>> x) {
    int n = sz(x), m = sz(x[0]);

    vector<pair<ll, int>> add;

    for (int i = 0; i < n; i++) {
        auto a = x[i];
        sort(a.begin(), a.end());
        for (int j = m-1; j >= m-k; j--) {
            ll x = a[j] + a[j-(m-k)];
            add.emplace_back(x, i);
        }
    }
    sort(add.rbegin(), add.rend());

    vector<int> cnt_plus(n);
    for (int i = 0; i < k * (n / 2); i++) {
        cnt_plus[add[i].second]++;
    }

    vector<vector<int>> neg(n), pos(n);
	for (int i = 0; i < n; i++) {
        vector<int> p(m); std::iota(p.begin(), p.end(), 0);
        sort(p.begin(), p.end(), [&](int a, int b) { return x[i][a] < x[i][b]; });
        int me = cnt_plus[i];
        for (int j = 0; j < k-me; j++) {
            neg[i].push_back(p[j]);
        }
        for (int j = 0; j < me; j++) {
            pos[i].push_back(p[m-j-1]);
        }
	}

    vector<vector<int>> ans(n, vector<int>(m, -1));
    for (int r = 0; r < k; r++) {
        vector<int> has(n, -1); // -1 = none, 1 = pos, 2 = neg
        int need = n / 2;
        for (int i = 0; i < n; i++) {
            if (!sz(pos[i]) && need && has[i] == -1) {
                has[i] = 2;
                need--;
            }
        }
        for (int i = 0; i < n; i++) {
            if (sz(neg[i]) && need && has[i] == -1) {
                has[i] = 2;
                need--;
            }
        }
        assert(need == 0);
        need = n / 2;
        for (int i = 0; i < n; i++) {
            if (sz(pos[i]) && need && has[i] == -1) {
                has[i] = 1;
            }
        }
        for (int i = 0; i < n; i++) {
            if (has[i] == 1) { // pos
                ans[i][pos[i].back()] = r;
                pos[i].pop_back();
            } else if (has[i] == 2) { // neg
                ans[i][neg[i].back()] = r;
                neg[i].pop_back();
            } else assert(false);
        }
    }

	allocate_tickets(ans);
	return cost(k, x, ans);
}

• Subtask 1: 11 / 11
• Subtask 2: 16 / 16
• Subtask 3: 14 / 14
• Subtask 4: 14 / 14
• Subtask 5: 12 / 12
• Subtask 6: 23 / 23
• Subtask 7: 10 / 10