이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <algorithm>
#include "tickets.h"
#include <vector>
#define ll long long
using namespace std;
const ll inf = 1e18 + 69;
#define pii pair<int, int>
vector<vector<int>> answer;
ll ans;
vector<pii> wtf[1502];
long long find_maximum(int k, vector<vector<int>> x) {
int n = x.size();
int m = x[0].size();
answer.resize(n);
for (int i = 0; i < n; i++) {
answer[i].resize(m, -1);
}
if (k == 1) {
vector<pii> d(n);
for (int i = 0; i < n; i++) {
d[i] = pii(x[i][m - 1] + x[i][0], i);
ans += x[i][m - 1];
}
sort(d.begin(), d.end());
vector<int> v(n);
for (int i = 0; i < n / 2; i++) {
int ind = d[i].second;
answer[ind][0] = 0;
ans -= d[i].first;
}
for (int i = n / 2; i < n; i++) {
int ind = d[i].second;
answer[ind][m - 1] = 0;
}
}
else if (k == m) {
vector<pii> all;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
all.push_back(pii(x[i][j], i * m + j));
}
}
sort(all.begin(), all.end());
for (int i = n * m / 2; i < n * m; i++) {
ans += all[i].first;
}
for (int i = 0; i < n * m / 2; i++) {
ans -= all[i].first;
}
pii mid = all[n + m / 2];
vector<int> ty(n); // -1 ++, -2 --, tver@ +- u vortex a sksvum +
int a = 0, b = 0;
vector<int> pm;
for (int i = 0; i < n; i++) {
if (pii(x[i][0], i * m) >= mid) {
ty[i] = -1;
a++;
}
else if (pii(x[i][m - 1], i * m + m - 1) < mid) {
ty[i] = -2;
b++;
}
else {
for (int j = 0; j < m; j++) {
wtf[i].push_back(pii(x[i][j], i * m + j));
}
ty[i] = lower_bound(wtf[i].begin(), wtf[i].end(), mid) - wtf[i].begin();
pm.push_back(i);
}
}
for (int i = 0; i < n; i++) {
if (ty[i] < 0) {
for (int j = 0; j < m; j++) {
answer[i][j] = j;
}
}
}
vector<int> zn(n), zv(n);
fill(zv.begin(), zv.end(), m - 1);
for (int j = 0; j < m; j++) {
int aa = n / 2 - a, bb = n / 2 - b;
for (int i = 0; i < n; i++) {
if (ty[i] < 0) continue;
if (zn[i] < ty[i] && aa) {
answer[i][zn[i]] = j;
zn[i]++;
aa--;
}
else {
answer[i][zv[i]] = j;
zv[i]--;
bb--;
}
}
}
}
allocate_tickets(answer);
return 69420;
}
/*
2 3 2
0 2 5
1 1 3
*/
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