(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^3log 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
", int ( 1ll * tree * F[0][n - 1] % p ) );
}
return 0;
}