[ABC302C] Almost Equal

由于 $n\leq 8$，所以可以考虑 $O(n!)$ 暴力枚举全排列，并且依次计算是否满足题意要求。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <cstdlib>
#include <string>
using namespace std;

#define ll long long

const int N = 2e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, m, t;
string s[9];

int cntdif(string x, string y)
{
	int cnt = 0;
	for (int i = 0; i < m; i++)
	{
		if (x[i] != y[i]) cnt++;
	}
	return cnt;
}

int main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
	{
		cin >> s[i];
	}
	sort(s + 1, s + n + 1);
	do
	{
		bool f = 1;
		for (int i = 1; i < n; i++)
		{
			if (cntdif(s[i], s[i + 1]) != 1)
			{
				f = 0;
				break;
			}
		}
		if (f)
		{
			printf("Yes\n");
			return 0;
		}
	} while (next_permutation(s + 1, s + n + 1));
	printf("No\n");
	return 0;
}