【SHOI2016】黑暗前的幻想乡

题面

题解

如果没有建筑公司的限制，那么就是个\(\mathrm{Matrix\;tree}\)板子

其实有了也一样

发现\(n\leq 17\)，考虑容斥

每次钦定一些建筑公司，计算它们包含的边的生成树的方案数

复杂度\(\mathrm{O}(2^nn^3)\)

代码

#include<cstdio>

#include<cstring>

#include<cctype>

#include<algorithm>

#define RG register

#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)

#define clear(x, y) memset(x, y, sizeof(x))

inline int read()

{

	int data = 0, w = 1; char ch = getchar();

	while(ch != '-' && (!isdigit(ch))) ch = getchar();

	if(ch == '-') w = -1, ch = getchar();

	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();

	return data * w;

}

const int N(20), Mod(1e9 + 7);

int n, a[N][N], m[N], popcnt[1 << 17], ans;

int U[N][N * N], V[N][N * N];

int solve(int S)

{

	for(RG int i = 1; i <= n; i++)

		for(RG int j = 1; j <= n; j++)

			a[i][j] = 0;

	for(RG int i = 1; i < n; i++)

		if(S & (1 << (i - 1))) for(RG int j = 1; j <= m[i]; j++)

		{

			int x = U[i][j], y = V[i][j];

			++a[x][x], ++a[y][y], --a[y][x], --a[x][y];

		}

	for(RG int i = 1; i <= n; i++)

		for(RG int j = 1; j <= n; j++)

			a[i][j] = (a[i][j] + Mod) % Mod;

	int ans = 1;

	for(RG int i = 2; i <= n; i++)

	{

		for(RG int j = i + 1; j <= n; j++)

			while(a[j][i])

			{

				int t = a[i][i] / a[j][i];

				for(RG int k = i; k <= n; k++)

					a[i][k] = (a[i][k] - 1ll * a[j][k] * t % Mod + Mod) % Mod,

						std::swap(a[i][k], a[j][k]);

				ans = Mod - ans;

			}

		ans = 1ll * ans * a[i][i] % Mod;

	}

	return ans;

}

int main()

{

#ifndef ONLINE_JUDGE

	file(cpp);

#endif

	n = read();

	for(RG int i = 1; i < n; i++)

	{

		m[i] = read();

		for(RG int j = 1; j <= m[i]; j++)

			U[i][j] = read(), V[i][j] = read();

	}

	for(RG int i = 0; i < 1 << (n - 1); i++)

		popcnt[i] = popcnt[i >> 1] + (i & 1);

	for(RG int i = 0; i < 1 << (n - 1); i++)

		ans = (ans + (((n - 1 - popcnt[i]) & 1) ?

					Mod - solve(i) : solve(i))) % Mod;

	printf("%d\n", ans);

	return 0;

}