Timus Online Judge1627--- Join

时间：2015-01-30 09:16:22 收藏：0 阅读：166

1627. Join

Time limit: 4.0 second
Memory limit: 64 MB

Businessman Petya recently bought a new house. This house has one floor withn ×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. Nextn lines contain exactlym 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 10⁹.

Samples

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

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

Tags: ( )

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: zchao1995@gmail.com 
    > 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]);
                                        }
<pre name="code" class="cpp">                                        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;}