[北京省选集训2019]生成树计数

题目

\((\sum_{i=1}^nw_i)^k\)直观理解一下就是

\[\sum_{\sum c_i=k}w_i^{c_i} \]

对于某种\(\{c_i\}\)，其出现次数就是\(\frac{k!}{\prod c_i!}\)

于是上面的式子就是

\[k!\sum_{c}\frac{w_i^{c_i}}{c_i!} \]

于是可以套路地构造一个\(\rm EGF\)，第\(i\)条边的\(\rm EGF\)就是\(\sum_{i=0}^\infty \frac{w_i^i}{i!}x^i=e^{w_ix}\)

于是我们要求的应该是\(k!\sum_T[k]\prod e^{w_ix}\)，不难发现这是一个矩阵树定理的形式，构造一个基尔霍夫矩阵消元求行列式就好了，唯一特别的就是行列式里面变成了多项式

我们可以\(O(k^2)\)的实现多项式乘法和乘法逆，于是复杂度就是\(O(n^3k^2)\)常数相当小

代码

#include<bits/stdc++.h>
#define re register
const int mod=998244353;
inline int dqm(int x) {return x<0?x+mod:x;}
inline int qm(int x) {return x>=mod?x-mod:x;}
struct Poly {int f[31];}a[31][31];
int n,k,fac[33],ifac[33],w[33][33];
inline int ksm(int a,int b) {
	int S=1;for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)S=1ll*S*a%mod;return S;
}
inline Poly operator+(const Poly &A,const Poly &B) {
	Poly C;for(re int i=0;i<=k;i++) C.f[i]=qm(A.f[i]+B.f[i]);return C;
}
inline Poly operator-(const Poly &A,const Poly &B) {
	Poly C;for(re int i=0;i<=k;i++) C.f[i]=dqm(A.f[i]-B.f[i]);return C;
}
inline Poly operator*(const Poly &A,const Poly &B) {
	Poly C;
	for(re int i=0;i<=k;i++) {
		C.f[i]=0;
		for(re int j=0;j<=i;j++) C.f[i]=qm(C.f[i]+1ll*A.f[j]*B.f[i-j]%mod);
	}
	return C;
}
inline Poly INV(const Poly &A) {
	Poly C;for(re int i=0;i<=k;i++) C.f[i]=0;
	int Inv=ksm(A.f[0],mod-2);C.f[0]=Inv;
	for(re int i=1;i<=k;i++) {
		for(re int j=1;j<=i;j++) C.f[i]=dqm(C.f[i]-1ll*A.f[j]*C.f[i-j]%mod);
		C.f[i]=1ll*C.f[i]*Inv%mod;
	}
	return C;
}
inline Poly get(int v) {
	Poly C;int nw=1;
	for(re int i=0;i<=k;i++) C.f[i]=1ll*nw*ifac[i]%mod,nw=1ll*nw*v%mod;
	return C;
}
int main() {
	scanf("%d%d",&n,&k);if(!k) return printf("%d\n",n>1?ksm(n,n-2):1),0;
	for(re int i=1;i<=n;i++)
		for(re int j=1;j<=n;j++) scanf("%d",&w[i][j]);
	fac[0]=ifac[0]=1;for(re int i=1;i<=k;i++) fac[i]=1ll*fac[i-1]*i%mod;
	ifac[k]=ksm(fac[k],mod-2);for(re int i=k-1;i;i--) ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
	for(re int i=1;i<=n;i++) 
		for(re int j=1;j<i;++j) {
			Poly t=get(w[i][j]);
			a[i][i]=a[i][i]+t;a[j][j]=a[j][j]+t;
			a[i][j]=a[i][j]-t;a[j][i]=a[j][i]-t;
		}
	int o=0;
	for(re int i=1;i<n;i++) {
		int p=i;
		for(re int j=i;j<n;++j) if(a[i][j].f[0]) {p=j;break;}
		if(p!=i) std::swap(a[i],a[p]),o^=1;
		Poly Inv=INV(a[i][i]);
		for(re int j=i+1;j<n;j++) {
			Poly t=Inv*a[j][i];
			for(re int k=i;k<n;++k)
				a[j][k]=a[j][k]-(t*a[i][k]);
		}
	}
	Poly ans;for(re int i=1;i<=k;i++) ans.f[i]=0;ans.f[0]=1;
	for(re int i=1;i<n;i++) ans=ans*a[i][i];
	if(!o) printf("%d\n",1ll*fac[k]*ans.f[k]%mod);
	else printf("%d\n",1ll*dqm(-fac[k])*ans.f[k]%mod);
	return 0;
}