Solution -「SHOI2016」「洛谷 P4336」黑暗前的幻想乡

(mathcal{Description})

  link.
  有一个 (n) 个结点的无向图，给定 (n-1) 组边集，求从每组边集选出恰一条边最终构成树的方案树。对 (10^9+7) 取模。
  (2le nle17)，边集大小 (0le m_ilefrac{n(n-1)}2)。

(mathcal{Solution})

  (n) 很小，考虑容斥。枚举这 (n-1) 个边集的子集，将子集内的边集的边加入图，用矩阵树定理求出生成树个数，容斥一下就好啦。复杂度 (mathcal O(2^nn^3))。

(mathcal{Code})

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>

const int MAXN = 17, MOD = 1e9 + 7;
int n, m, d[MAXN + 5][MAXN + 5];
std::vector<std::pair<int, int> > able[MAXN + 5]; 

inline int qkpow ( int a, int b, const int p = MOD ) {
	int ret = 1;
	for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = 1ll * ret * ( b & 1 ? a : 1 ) % p;
	return ret;
}

inline int det ( int d[MAXN + 5][MAXN + 5] ) {
	int ret = 1, swp = 1;
	for ( int i = 1; i < n; ++ i ) {
		for ( int j = i; j < n; ++ j ) {
			if ( d[j][i] ) {
				if ( i ^ j ) std::swap ( d[i], d[j] ), swp *= -1;
				break;
			}
		}
		if ( ! d[i][i] ) return 0;
		ret = 1ll * ret * d[i][i] % MOD;
		int inv = qkpow ( d[i][i], MOD - 2 );
		for ( int j = i + 1; j < n; ++ j ) {
			int c = 1ll * inv * d[j][i] % MOD;
			for ( int k = i; k < n; ++ k ) d[j][k] = ( d[j][k] - 1ll * c * d[i][k] % MOD + MOD ) % MOD;
		}
	}
	return ( ret * swp + MOD ) % MOD;
}

int main () {
	scanf ( "%d", &n );
	for ( int i = 1, m; i < n; ++ i ) {
		scanf ( "%d", &m );
		for ( int u, v; m --; ) {
			scanf ( "%d %d", &u, &v );
			able[i].push_back ( { u, v } );
		}
	}
	int ans = 0;
	for ( int s = 1; s < 1 << n >> 1; ++ s ) {
		int bit = 0; memset ( d, 0, sizeof d );
		for ( int i = 1; i < n; ++ i ) {
			if ( ( s >> i - 1 ) & 1 ) {
				++ bit;
				for ( int j = 0; j ^ able[i].size (); ++ j ) {
					int u = able[i][j].first, v = able[i][j].second;
					++ d[u][u], ++ d[v][v], -- d[u][v], -- d[v][u];
					if ( d[u][v] < 0 ) d[u][v] += MOD;
					if ( d[v][u] < 0 ) d[v][u] += MOD;
				}
			}
		}
		ans = ( ans + ( ( bit & 1 ) ^ ( n & 1 ) ? det ( d ) : -det ( d ) ) ) % MOD;
	}
	printf ( "%d
", ( ans + MOD ) % MOD );
	return 0;
}