#include "tickets.h"
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long LL;
typedef vector<int> IVec;
typedef pair<LL, LL> pii;
long long find_maximum(int k, vector<IVec> d) {
int N = d.size(), M = d[0].size(); // N color ×M tickets
vector<IVec> o(N); // o.resize(N);
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) o[i].push_back(j); // o[N][i]:index of ticket which is ith min from left
// // Sort each color's tickets by their num
sort(o[i].begin(), o[i].end(), [&](int a, int b) { return d[i][a] < d[i][b]; });
}
auto dval = [&](int c, int i) { return d[c][o[c][i]]; }; // num of ticket of color c which is ith form left
LL cost = 0;
IVec Plus(N, 0); // Plus[c] => How many '+'s the color c will send
priority_queue<pii> gain; // gain: (change to cost by turning a - into a +, color index) TODO: ?
for (int c = 0; c < N; c++) {
for (int j = 0; j < k; j++) cost -= dval(c, j); // First incur cost of all -s
gain.push(make_pair(dval(c, M - 1 - Plus[c]) + dval(c, k - 1 - Plus[c]), c)); // Queue a possible - to +
}
for (int i = 0; i < N * k / 2; i++) { // Take ck/2 +s
pii it = gain.top(); gain.pop();
cost += it.first;
int c = it.second; // If the color hasnt reached k +s, queue another possible +
if (++Plus[c] < k)
gain.push(pii(dval(c, M - 1 - Plus[c]) + dval(c, k - 1 - Plus[c]), c));
}
while (!gain.empty()) gain.pop(); // Empty out the priority queue, will use again later
for (int i = 0; i < N; i++) gain.push(pii(Plus[i], i)); // gain: (+s remaining, color index)
// minus_count[color] => How many -s the color has sent so far
vector<int> minus_count;
minus_count.resize(N, 0);
// Prepare answer vector dimensions and fill with -1s
vector<IVec> answer(N, IVec(M, -1));
IVec take(N, 0); // take[color] => Does this color send a + in this days
for (int i = 0; i < k; i++) { // Simulate k days of tickets
for (int j = 0; j < N / 2; j ++) // Pick N/2 companies with the most +s left
take[gain.top().second] = 1, gain.pop();
// Check if each color sent a - or +, and change answer accordingly
for (int j = 0; j < N; j++) {
if (take[j]) {
// Company j sent a + this round, send the smallest +
// Queue the color back
answer[j][o[j][M - Plus[j]]] = i;
Plus[j]--;
gain.push(pii(Plus[j], j));
} else {
// Company j sent a - this round, send the smallest -
answer[j][o[j][minus_count[j]]] = i;
minus_count[j]++;
}
take[j] = 0;
}
}
// Return to grader
allocate_tickets(answer);
return cost;
}
