题意太长，就放里面了

CF578F

Description

19-2-16-1

Constraints

n,m≤100n,m\le 100，*的个数不超过300300

Solution

神仙结论题

对盒子中所有的顶点（一共有(n+1)(m+1)(n + 1)(m + 1)个顶点）进行黑白染色，,如果把一面镜子看成一条边的话，那么一面镜子连接了两个颜色相同的顶点。

结论：完美盒子的充要条件是某种颜色的点形成了一棵树。

接下来是十you分chou优you美chang的证明：

19-2-16-2

枚举是白点形成了树还是黑点形成了树，然后把已知的边连上，就可以把原来的问题规模缩小到与*的个数同规模的生成树计数了，直接Matrix−TreeMatrix-Tree，时间复杂度O(3003)O(300^3)

注意：若形成的联通块数量大于300，则一定无解

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 100 + 10;

int Mod;

int N, M, Id[Maxn][Maxn];
char S[Maxn][Maxn];

inline int Pow (int a, int b)
{
	int ans = 1;
	for (int i = b; i; i >>= 1, a = (LL)a * a % Mod) if (i & 1) ans = (LL)ans * a % Mod;
	return ans;
}

inline void Add (int &a, int b) { a += b; if (a >= Mod) a -= Mod; }

namespace DSU
{
	const int maxn = Maxn * Maxn;
	int fa[maxn];

	inline void init (int maxn) { for (int i = 0; i <= maxn; ++i) fa[i] = i; }
	inline int find (int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
	inline int judge (int x, int y) { return find(x) == find(y); }
	inline void link (int x, int y) { fa[find(x)] = find(y); }
}

namespace Matrix
{
	const int maxn = 300 + 10;

	int A[maxn][maxn];

	inline void init () { memset(A, 0, sizeof A); }

	inline void add_edge (int x, int y) { ++A[x][x], ++A[y][y], --A[x][y], --A[y][x]; }

	inline int gauss (int n)
	{
		int ans = 1;
		for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) A[i][j] = (A[i][j] + Mod) % Mod;

		for (int i = 1; i <= n; ++i)
		{
			if (!A[i][i])
			{
				int id = i;
				for (int j = i + 1; j <= n; ++j)
					if (A[j][i])
					{
						swap(A[i], A[j]);
						ans = (LL)ans * (Mod - 1) % Mod;
						break;
					}
				if (!ans) return 0;
			}

			int inv = Pow(A[i][i], Mod - 2);
			for (int j = i + 1; j <= n; ++j)
			{
				int now = (LL)A[j][i] * inv % Mod;
				for (int k = i + 1; k <= n; ++k) Add (A[j][k], Mod - (LL)now * A[i][k] % Mod);
				A[j][i] = 0;
			}
		}

		for (int i = 1; i <= n; ++i) ans = (LL)ans * A[i][i] % Mod;

		return ans;
	}
}

int P[Maxn * Maxn], ans;

inline void Solve ()
{
	DSU :: init ((N + 1) * (M + 1));
	for (int i = 1; i <= N; ++i)
		for (int j = 1; j <= M; ++j)
		{
			if (S[i][j] == '*') continue;
			int x, y;
			if (S[i][j] == '/') x = Id[i][j + 1], y = Id[i + 1][j];
			else x = Id[i][j], y = Id[i + 1][j + 1];

			if (DSU :: judge (x, y)) { puts("0"); return ; }
			DSU :: link (x, y);
		}

	for (int k = 0; k < 2; ++k)
	{

		int cnt = 0;
		Matrix :: init ();

		for (int i = 1; i <= N + 1; ++i)
			for (int j = 1; j <= M + 1; ++j)
				if ((i + j) % 2 == k && DSU :: find(Id[i][j]) == Id[i][j]) P[Id[i][j]] = ++cnt;

		if (cnt > 300) continue;

		for (int i = 1; i <= N; ++i)
			for (int j = 1; j <= M; ++j)
				if (S[i][j] == '*')
				{
					int x, y;
					if ((i + j) % 2 == k) x = Id[i][j], y = Id[i + 1][j + 1];
					else x = Id[i][j + 1], y = Id[i + 1][j];
					x = P[DSU :: find(x)], y = P[DSU :: find(y)];
					Matrix :: add_edge (x, y);
				}

		Add (ans, Matrix :: gauss (cnt - 1));
	}

	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>(), M = read<int>(), Mod = read<int>();
	for (int i = 1; i <= N; ++i) scanf("%s", S[i] + 1);
}

inline void Init ()
{
	int tot_cnt = 0;
	for (int i = 1; i <= N + 1; ++i)
		for (int j = 1; j <= M + 1; ++j)
			Id[i][j] = ++tot_cnt;
}

int main()
{

#ifndef ONLINE_JUDGE
	freopen("xxx.in", "r", stdin);
	freopen("xxx.out", "w", stdout);
#endif

	Input();
	Init();
	Solve();

	return 0;
}