CF1450C2 Errich-Tac-Toe (Hard Version) 题解

还比较好想。

考虑对棋盘三染色，每连着的三个必然有三种颜色。只需要将三种颜色中的其中两种分别染成 O 和 X 或 X 和 O 即可。

总共 $6$ 种染法，选最小的那个做，鸽巢原理容易发现操作次数不超过其限制。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

const int N = 305;

int col[N][N];
char c[N][N];
int t, n, sum = 0;

char res[N][N], tmp[N][N];

void solve(int c1, int c2, char f1, char f2)
{
	int cnt = 0;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			tmp[i][j] = c[i][j];
			if (c[i][j] == '.') continue;
			if (col[i][j] == c1)
			{
				if (c[i][j] != f1)
				{
					cnt++;
					tmp[i][j] = f1;
				}
			}
			else if (col[i][j] == c2)
			{
				if (c[i][j] != f2)
				{
					cnt++;
					tmp[i][j] = f2;
				}
			}
		}
	}
	if (cnt <= sum / 3)
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				res[i][j] = tmp[i][j];
			}
		}
	}
}

int main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin >> t;
	while (t--)
	{
		cin >> n;
		sum = 0;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++) cin >> c[i][j], col[i][j] = (i + j) % 3, sum += (c[i][j] != '.');
		}
		solve(0, 1, 'O', 'X');
		solve(0, 1, 'X', 'O');
		solve(1, 2, 'O', 'X');
		solve(1, 2, 'X', 'O');
		solve(2, 0, 'O', 'X');
		solve(2, 0, 'X', 'O');
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++) cout << res[i][j];
			cout << "\n";
		}
	}

	return 0;
}