自然地想到容斥原理

然後套個矩陣樹就行了

求行列式的時候只有換行要改變符號啊QAQ

複雜度為\(O(2^n * n^3)\)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 20;
const int mod = 1e9 + 7;
inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline void dec(int &a, int b) { a -= b; if(a < 0) a += mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline int inv(int a) {
    int ret = 1;
    for(int k = mod - 2; k; k >>= 1, a = mul(a, a))
        if(k & 1) ret = mul(ret, a);
    return ret;
}
    
int N, ans;
int M[sid], G[sid][sid];
struct edge { int u, v; } E[sid][400];

inline int Guass(int n) {
    int sign = 1;
    rep(i, 1, n) {
        int pos = i;
        rep(j, i + 1, n) if(G[j][i]) pos = j;
        swap(G[i], G[pos]); 
        if(i != pos) sign *= -1;
        if(!G[i][i]) return 0;
        int Inv = inv(G[i][i]);
        rep(j, i + 1, n) {
            int t = mul(G[j][i], Inv);
            rep(k, i, n) dec(G[j][k], mul(G[i][k], t));
        }
    }
    int ret = 1;
    rep(i, 1, n) ret = mul(ret, G[i][i]);
    if(sign == 1) return ret;
    else return mod - ret;
}

inline int calc(int S) {
    memset(G, 0, sizeof(G));
    rep(i, 1, N - 1) {
        if(!(S & (1 << i - 1))) continue;
        rep(j, 1, M[i]) {
            int u = E[i][j].u, v = E[i][j].v;
            inc(G[u][u], 1); inc(G[v][v], 1);
            dec(G[u][v], 1); dec(G[v][u], 1);
        }
    }
    return Guass(N - 1);
}

inline void dfs(int o, int S, int num) {
    if(o == N) {
        if((N - 1 - num) & 1) dec(ans, calc(S));
        else inc(ans, calc(S));
        return;
    }
    dfs(o + 1, S, num);
    dfs(o + 1, S | (1 << o - 1), num + 1);
}
    
int main() {
    freopen("pp.in", "r", stdin);
    cin >> N;
    rep(i, 1, N - 1) {
        cin >> M[i];
        rep(j, 1, M[i]) 
        cin >> E[i][j].u >> E[i][j].v;
    }
    dfs(1, 0, 0);
    printf("%d\n", ans);
    return 0;
}