이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "tickets.h"
#include <bits/stdc++.h>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
using namespace std;
using lint = long long;
using llf = long double;
using pi = pair<lint, lint>;
const int MAXN = 1505;
struct node{
int x, y, v;
bool operator<(const node &n)const{
return v < n.v;
}
};
void FUCKING_BACKTRACK(vector<vector<int>> B, int k){
deque<int> dq[MAXN];
int bcnt[MAXN] = {};
for(int i=0; i<sz(B); i++){
for(auto &j : B[i]){
if(j > 0) bcnt[i]++;
}
for(int j=0; j<sz(B[i]); j++){
if(B[i][j] == -1) dq[i].push_back(j);
}
for(int j=0; j<sz(B[i]); j++){
if(B[i][j] == +1) dq[i].push_back(j);
}
for(int j=0; j<sz(B[i]); j++){
if(B[i][j] == 0) B[i][j] = -1;
else B[i][j] = 0;
}
}
for(int i = 0; i < k; i++){
vector<int> idx(sz(B));
iota(all(idx), 0);
sort(all(idx), [&](const int &a, const int &b){
return bcnt[a] < bcnt[b];
});
for(int j=0; j<sz(B); j++){
int x = idx[j];
int y = (j < sz(B) / 2 ? dq[x].front() : dq[x].back());
B[x][y] = i;
if(j < sz(B) / 2) dq[x].pop_front();
else{
dq[x].pop_back();
bcnt[x]--;
}
}
}
allocate_tickets(B);
}
lint dp[1500][10000];
int track[1500][10000];
long long find_maximum(int k, std::vector<std::vector<int>> x) {
int n = sz(x);
int m = sz(x[0]);
vector<vector<int>> B(n);
for(auto &i : B) i.resize(m);
if(max(n, m) > 80) return 69;
vector<node> v;
fill(dp[0] + 1, dp[0] + 10000, -1e18);
for(int i=0; i<n; i++){
fill(dp[i + 1], dp[i + 1] + 10000, -1e18);
lint sum = 0;
for(int j=0; j<k; j++){
sum -= x[i][j];
}
for(int j = 0; j <= k; j++){
for(int x = j; x <= (i + 1) * k; x++){
if(dp[i + 1][x] < dp[i][x - j] + sum){
dp[i + 1][x] = dp[i][x - j] + sum;
track[i + 1][x] = j;
}
}
if(j != m) sum += x[i][m - 1 - j];
if(j != k) sum += x[i][k - 1 - j];
}
}
int fuck = (n / 2) * k;
for(int i = n; i; i--){
int nxt = track[i][fuck];
fuck -= nxt;
for(int j = 0; j < k - nxt; j++) B[i-1][j] = -1;
for(int j = m - nxt; j < m; j++) B[i-1][j] = +1;
}
/*
sort(all(v));
int l = 0, r = sz(v) - 1;
vector<int> cnt(n);
lint ret = 0;
for(int i=0; i<(n/2)*k; i++){
while(cnt[v[l].x] == k) l++;
B[v[l].x][v[l].y] = -1;
ret -= v[l].v;
cnt[v[l].x]++;
l++;
while(cnt[v[r].x] == k) r--;
B[v[r].x][v[r].y] = 1;
ret += v[r].v;
cnt[v[r].x]++;
r--;
}*/
FUCKING_BACKTRACK(B, k);
return dp[n][(n/2)*k];
}
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