Solution -「HDU 5498」Tree

\(\mathcal{Description}\)

  link.

  给定一个 \(n\) 个结点 \(m\) 条边的无向图，\(q\) 次操作每次随机选出一条边。问 \(q\) 条边去重后构成生成树的方案总数，对 \(p\) 取模。

\(\mathcal{Solution}\)

  首先求出 \(n-1\) 条边构成生成树的方案数，显然矩阵树定理。

  接着，令 \(f(i,j)\) 表示操作 \(i\) 次，去重后有 \(j\) 条边的方案数。那么有：

\[f(i,j)=jf(i-1,j)+(m-j+1)f(i-1,j-1)
\]

  这个式子可以矩阵快速幂优化，最后把上面两个东西乘起来就是总方案啦。复杂度 \(\operatorname{O}(n^3\log q)\)。

\(\mathcal{Code}\)

  这个 HDU 它一直 SF 呢 qwq。理论 AC 代码如下 w。

#include <cstdio>
#include <cstring>
#include <assert.h>
#include <iostream>

const int MAXN = 100;
int n, m, p, q, K[MAXN + 5][MAXN + 5];

inline void add ( const int u, const int v ) {
	++ K[u][u], ++ K[v][v], -- K[u][v], -- K[v][u];
	if ( K[u][v] < 0 ) K[u][v] += p;
	if ( K[v][u] < 0 ) K[v][u] += p;
}

inline int det ( const int n ) {
	int ret = 1, swp = 1;
	for ( int i = 1; i < n; ++ i ) {
		for ( int j = i + 1; j < n; ++ j ) {
			for ( ; K[j][i]; std::swap ( K[i], K[j] ), swp *= -1 ) {
				int d = K[i][i] / K[j][i];
				for ( int k = i; k < n; ++ k ) K[i][k] = ( K[i][k] - 1ll * d * K[j][k] % p + p ) % p;
			}
		}
		if ( ! ( ret = 1ll * ret * K[i][i] % p ) ) return 0;
	}
	return ( ret * swp + p ) % p;
}

struct Matrix {
	int n, m, mat[MAXN + 5][MAXN + 5];
	Matrix () {} Matrix ( const int tn, const int tm ): n ( tn ), m ( tm ), mat {} {}
	inline int* operator [] ( const int key ) { return mat[key]; }
	inline Matrix operator * ( Matrix t ) {
		assert ( m == t.n );
		Matrix ret ( n, t.m );
		for ( int i = 0; i <= n; ++ i ) {
			for ( int k = 0; k <= m; ++ k ) {
				for ( int j = 0; j <= t.m; ++ j ) {
					ret[i][j] = ( ret[i][j] + 1ll * mat[i][k] * t[k][j] ) % p;
				}
			}
		}
		return ret;
	}
};

inline Matrix qkpow ( Matrix A, int b ) {
	Matrix ret ( A.n, A.m );
	for ( int i = 0; i <= A.n; ++ i ) ret[i][i] = 1;
	for ( ; b; A = A * A, b >>= 1 ) if ( b & 1 ) ret = ret * A;
	return ret;
}

int main () {
	int T;
	for ( scanf ( "%d", &T ); T --; memset ( K, 0, sizeof K ) ) {
		scanf ( "%d %d %d %d", &n, &m, &p, &q );
		for ( int i = 1, u, v; i <= m; ++ i ) scanf ( "%d %d", &u, &v ), add ( u, v );
		int tree = det ( n );
		if ( ! tree || q < n - 1 ) { puts ( "0" ); continue; }
		Matrix F ( 0, n - 1 ), T ( n - 1, n - 1 );
		F[0][0] = 1;
		for ( int i = 0; i < n; ++ i ) {
			T[i][i] = i;
			if ( i ) T[i - 1][i] = n - i;
		}
		F = F * qkpow ( T, q );
		printf ( "%d\n", int ( 1ll * tree * F[0][n - 1] % p ) );
	}
	return 0;
}