【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;
}