returns (score, finished)

returns (score, finished)

998244353

returns char printed OR info_nothing_printed OR error

returns char printed OR info_nothing_printed OR error

obliczanie wyniku przy znanej liczbie inwersji w kazdym wezle

uses global array: int priority[n_streets];

uses global array: int priority[n_streets];

void gen_ones() { for(int road : ins) { if(!street_freq[road]) { continue; } int len = 1; cycle.emplace_back(road, len); cycle_sum += len; } }

void gen_ones() { for(int road : ins) { if(!street_freq[road]) { continue; } int len = 1; cycle.emplace_back(road, len); cycle_sum += len; } }

1) Looking for 'target' in a sorted array.

O(2^N) f(next_index, sum_of_first_set, sum_of_second_set) f(next_index, sumFirst - sumSecond)

<<^^<>>^v^<v^vv<>v

<<^^<>>^v^<v^vv<>v

vector<pair<int,int>> directions{{-2,-1},{-2,1},{-1,2},{-1,-2},{1,2},{1,-2},{2,-1},{2,1}};

O(N^2 log(N)) solution first, then O(N^2) below

O(N! * N) but way faster in practice because of diagonals