Timus Online Judge1627--- Join

1627. Join

Time limit: 4.0 second
Memory limit: 64 MB

Businessman Petya recently bought a new house. This house has one floor with n × m square rooms, placed in rectangular lattice. Some rooms are pantries and the other ones are bedrooms. Now he wants to join all bedrooms with doors in such a way that there will be exactly one way between any pair of them. He can make doors only between neighbouring bedrooms (i.e. bedrooms having a common wall). Now he wants to count the number of different ways he can do it.

Input

First line contains two integers n and m (1 ≤ n, m ≤ 9)  — the number of lines and columns in the lattice. Next n lines contain exactly m characters representing house map, where "." means bedroom and "*" means pantry. It is guaranteed that there is at least one bedroom in the house.

Output

Output the number of ways to join bedrooms modulo 10 ⁹.

Samples

input	output
2 2 .. ..	4
2 2 *. .*	0

Problem Source: SPbSU ITMO contest. Petrozavodsk training camp. Winter 2008.

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Difficulty: 1154     Printable version     Submit solution     Discussion (13)
All submissions (1680)     All accepted submissions (428)     Solutions rating (223)

还是用拉普拉斯矩阵, 消元时用类似辗转相除的办法,保证中间过程为整数

/*************************************************************************
    > File Name: st10.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年01月29日 星期四 16时22分40秒
 ************************************************************************/

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

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
const int N = 210;
const int mod = 1000000000;

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
char str[20][20];
int ord[N];

LL mat[N][N];

LL Det (int n)
{
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			mat[i][j] %= mod;
		}
	}
	LL res = 1;
	for (int i = 0; i < n; ++i)
	{
		if (!mat[i][i])
		{
			bool flag = false;
			for (int j = i + 1; j < n; ++j)
			{
				if (mat[j][i])
				{
					flag = true;
					for (int k = i; k < n; ++k)
					{
						swap (mat[i][k], mat[j][k]);
					}
					res = -res;
					break;
				}
			}
			if (!flag)
			{
				return 0;
			}
		}
		for (int j = i + 1; j < n; ++j)
		{
			while (mat[j][i])
			{
				LL t = mat[i][i] / mat[j][i];
				for (int k = i; k < n; ++k)
				{
					mat[i][k] = (mat[i][k] - t * mat[j][k]) % mod;
					swap (mat[i][k], mat[j][k]);
				}
				res = -res;
			}
		}
		res = (res * mat[i][i]) % mod;
	}
	return (res + mod) % mod;
}

int main()
{
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		int cnt = 0;
		memset (mat, 0, sizeof(mat));
		for (int i = 0; i < n; ++i)
		{
			scanf("%s", str[i]);
			for (int j = 0; j < m; ++j)
			{
				if (str[i][j] == '.')
				{
					ord[i * m + j] = cnt++;
				}
			}
		}
		for (int i = 0; i < n; ++i)
		{
			for (int j = 0; j < m; ++j)
			{
				if (str[i][j] == '.')
				{
					for (int k = 0; k < 4; ++k)
					{
						int x = i + dir[k][0];
						int y = j + dir[k][1];
						if (x < 0 || x >= n || y < 0 || y >= m || str[x][y] != '.')
						{
							continue;
						}
						mat[ord[i * m + j]][ord[x * m + y]] = -1;
					}
				}
			}
		}
		for (int i = 0; i < cnt; ++i)
		{
			LL ret = 0;
			for (int j = 0; j < cnt; ++j)
			{
				if (i != j && mat[i][j])
				{
					++ret;
				}
			}
			mat[i][i] = ret;
		}
		LL ans = Det(cnt - 1);
		printf("%lld\n", ans);
	}
	return 0;
}