Submission #163221

#	Time	Username	Problem	Language	Result	Execution time	Memory
163221		qkxwsm	Sajam (COCI18_sajam)	C++14	90 / 90	103 ms	1444 KiB

sajam

#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define MAXN 1013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, K;
bitset<MAXN> grid[MAXN];
int freq[MAXN];

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	cout << fixed << setprecision(12);
	cerr << fixed << setprecision(4);
	cin >> N >> K;
	FOR(i, 0, N)
	{
		string temps; cin >> temps;
		FOR(j, 0, N)
		{
			grid[i][j] = (temps[j] == 'o');
		}
	}
	FOR(i, 0, N)
	{
		FOR(j, 0, N)
		{
			if (i == j) continue;
			grid[j] ^= grid[i];
		}
		grid[i].reset();
		int res = 0;
		FOR(j, 0, N)
		{
			int x = grid[j].count();
			res += min(x, N - x);
			// FOR(k, 0, N)
			// {
			// 	cerr << grid[j][k] << ' ';
			// }
			// cerr << endl;
		}
		if (res <= K)
		{
			cout << "DA\n";
			return 0;
		}
	}
	if (N == K)
	{
		if (N == 4)
		{
			//wtf stupid case.
			//try each of them as the changed guy.
			FOR(i, 0, 2)
			{
				//toggle row 1.
				FOR(j, 0, N) grid[0][j] = grid[0][j] ^ 1;
				FOR(j, 0, 2)
				{
					//toggle row 2.
					FOR(k, 0, N) grid[1][k] = grid[1][k] ^ 1;
					FOR(k, 0, 2)
					{
						//toggle row 3.
						//check if it's ok
						FOR(m, 0, N) grid[2][m] = grid[2][m] ^ 1;
						FOR(m, 0, N)
						{
							FOR(n, 0, N)
							{
								cerr << grid[m][n];
							}
							cerr << endl;
						}
						FOR(m, 0, N) freq[m] = 0;
						bool ok = true;
						FOR(m, 0, N - 1)
						{
							if (grid[m].count() != 2)
							{
								ok = false;
								break;
							}
							int x = grid[m]._Find_first(), y = grid[m]._Find_next(x);
							freq[x]++; freq[y]++;
						}
						if (!ok) continue;
						// cerr << "OK\n";
						sort(freq, freq + N);
						FOR(i, 0, N - 1)
						{
							if (freq[i] != 1)
							{
								ok = false;
								break;
							}
						}
						if (ok)
						{
							cout << "DA\n";
							return 0;
						}
						//ok now literally just check fi it works.
					}
				}
			}
			cout << "NE\n";
			return 0;
		}
		// FOR(j, 0, N)
		// {
		// 	FOR(k, 0, N)
		// 	{
		// 		cerr << grid[j][k];
		// 	}
		// 	cerr << endl;
		// }
		FOR(j, 0, N - 1)
		{
			if (grid[j].count() != 2)
			{
				FOR(k, 0, N) grid[j][k] = grid[j][k] ^ 1;
			}
			if (grid[j].count() != 2)
			{
				cout << "NE\n";
				return 0;
			}
			int x = grid[j]._Find_first(), y = grid[j]._Find_next(x);
			// cerr << x << " -> " << y << endl;
			freq[x]++; freq[y]++;
		}
		sort(freq, freq + N);
		FOR(i, 0, N - 1)
		{
			if (freq[i] != 1)
			{
				cout << "NE\n";
				return 0;
			}
		}
		cout << "DA\n";
		return 0;
	}
	//bleh now the N=K case
	cout << "NE\n";
	return 0;
	//there has to be a row with <=1.
	//fix a specific row and column.
	//What's the minimum we could have on after a series of operations.
}

• Subtask 1: 15 / 15
• Subtask 2: 15 / 15
• Subtask 3: 15 / 15
• Subtask 4: 15 / 15
• Subtask 5: 15 / 15
• Subtask 6: 15 / 15