#include "tickets.h"
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 300;
const int MAX_M = 300;
const int MAX_K = 300;
const long long INF = 1e15;
int x[MAX_N + 1][MAX_M + 1];
long long sx[MAX_N + 1][MAX_M + 1];
int tai[MAX_N + 1][MAX_N * MAX_K + 1];
long long dp[MAX_N + 1][MAX_N * MAX_K + 1];
vector<int> mins[MAX_N + 1], maxs[MAX_N + 1];
bool isMax[MAX_N + 1];
long long find_maximum( int k, vector<vector<int>> X ) {
int n = X.size();
int m = X[0].size();
vector<vector<int>> answer( n );
for ( int i = 1; i <= n; i++ ) {
answer[i - 1].resize( m );
for ( int j = 1; j <= m; j++ ) {
x[i][j] = X[i - 1][j - 1];
sx[i][j] = sx[i][j - 1] + x[i][j];
answer[i - 1][j - 1] = -1;
}
}
for ( int i = 1; i <= n; i++ ) {
for ( int a = 0; a <= k * i; a++ )
dp[i][a] = -INF;
for ( int a = 0; a <= k * (i - 1); a++ ) {
for ( int b = 0; b <= k; b++ ) {
if ( dp[i - 1][a] - sx[i][b] + sx[i][m] - sx[i][m - (k - b)] > dp[i][a + b] ) {
dp[i][a + b] = dp[i - 1][a] - sx[i][b] + sx[i][m] - sx[i][m - (k - b)];
tai[i][a + b] = b;
}
}
}
//for ( int a = 0; a <= k * i; a++ )
// printf( "(%d %d) ", dp[i][a], tai[i][a] );
//printf( "\n" );
}
int a = k * n / 2;
for ( int i = n; i >= 1; i-- ) {
int b = tai[i][a];
for ( int j = b; j >= 1; j-- )
mins[i].push_back( j );
for ( int j = m; j > m - (k - b); j-- )
maxs[i].push_back( j );
//printf( "ba %d %d\n", i, b );
a -= b;
}
for ( int pas = 0; pas < k; pas++ ) {;
for ( int i = 1; i <= n; i++ )
isMax[i] = false;
vector<pair<int, int>> maxx;
for ( int i = 1; i <= n; i++ ) {
if ( !maxs[i].empty() )
maxx.push_back( { (mins[i].empty() ? 0 : x[i][maxs[i].back()]), i } );
}
sort( maxx.begin(), maxx.end() );
for ( int i = 0; i < n / 2; i++ ) {
isMax[maxx[i].second] = true;
//printf( "%d\n", maxx[i].second );
answer[maxx[i].second - 1][maxs[maxx[i].second].back() - 1] = pas;
maxs[maxx[i].second].pop_back();
}
for ( int i = 1; i <= n; i++ ) {
if ( isMax[i] )
continue;
if ( mins[i].empty() )
exit( 1 );
answer[i - 1][mins[i].back() - 1] = pas;
mins[i].pop_back();
}
}
allocate_tickets( answer );
return dp[n][k * n / 2];
}
