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#include #include #include #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;}