HDU5279 YJC plays Minecraft (CDQ+NTT)

HDU-5279(CDQ+NTT)

本质其实是要求\(n\)个点森林数量\(dp_n\)，\(n\)个点森林并且\(1,n\)在同一连通块的数量\(f_n\)

总方案就是\(\Pi dp_{a_i}\cdot 2^n-\Pi f_{a_i}\)

就是减去所有环都连着，并且\(1,a_i\)连着的方案数

我们知道\(n\)个点树的数量是\(n^{n-2}\)(Prufer序列)

枚举\(1\)号点所在树的大小为\(j\)，则\(dp_i=dp_{i-j}\cdot j^{j-2}\cdot C(i-1,j-1)\)

可以看到是一个与差值有关的转移，可以通过\(CDQ+NTT\)优化

然后就是求\(f_i\)，其实就是枚举\(1,i\)号点所在树的大小为\(j\)，\(f_i=dp_{i-j}\cdot C(i-2,j-2) \cdot j^{j-2}\)，可以直接\(NTT\)解决

#include<bits/stdc++.h>
using namespace std;

#define reg register
typedef long long ll;
#define rep(i,a,b) for(reg int i=a,i##end=b;i<=i##end;++i)
#define drep(i,a,b) for(reg int i=a,i##end=b;i>=i##end;--i)

template <class T> inline void cmin(T &a,T b){ ((a>b)&&(a=b)); }
template <class T> inline void cmax(T &a,T b){ ((a<b)&&(a=b)); }

char IO;
int rd(){
	int s=0,f=0;
	while(!isdigit(IO=getchar())) if(IO=='-') f=1;
	do s=(s<<1)+(s<<3)+(IO^'0');
	while(isdigit(IO=getchar()));
	return f?-s:s;
}

const ll N=(1<<18)|4,P=998244353;

int n=1e5;
ll Fac[N],Inv[N];
ll Tr[N];
ll qpow(ll x,ll k){
	ll res=1;
	for(;k;k>>=1,x=x*x%P) if(k&1) res=res*x%P;
	return res;
}
ll dp[N],f[N],A[N],B[N];
int rev[N];

void NTT(int n,ll *a,int f) {
	rep(i,0,n-1) if(rev[i]<i) swap(a[i],a[rev[i]]);
	for(reg int i=1;i<n;i<<=1) {
		ll w=qpow(f==1?3:(P+1)/3,(P-1)/i/2);
		for(reg int l=0;l<n;l+=2*i) {
			ll e=1;
			for(reg int j=l;j<l+i;++j,e=e*w%P) {
				ll t=a[j+i]*e%P;
				a[j+i]=a[j]-t;
				a[j]=a[j]+t;
				(a[j+i]<0&&(a[j+i]+=P));
				(a[j]>=P&&(a[j]-=P));
			}
		}
	}
	if(f==-1) {
		ll base=qpow(n,P-2);
		rep(i,0,n-1) a[i]=a[i]*base%P;
	}
}

void Solve(int l,int r) {  // CDQ计算dp
	if(l==r) return;
	int mid=(l+r)>>1;
	Solve(l,mid);
	int R=1,cc=-1;
	while(R<=r-l+1) R<<=1,cc++;
	rep(i,1,R) rev[i]=(rev[i>>1]>>1)|((i&1)<<cc);
	rep(i,0,R) A[i]=B[i]=0;
	rep(i,l,mid) A[i-l]=dp[i]*Inv[i]%P;
	rep(i,1,r-l) B[i]=Inv[i-1]%P*Tr[i]%P;
	NTT(R,A,1),NTT(R,B,1);
	rep(i,0,R-1) A[i]=A[i]*B[i]%P;
	NTT(R,A,-1);
	rep(i,mid+1,r) dp[i]=(dp[i]+A[i-l]*Fac[i-1])%P;
	Solve(mid+1,r);
}

int main(){
	Fac[0]=Fac[1]=Inv[0]=Inv[1]=1;
	rep(i,2,N-1) {
		Fac[i]=Fac[i-1]*i%P;
		Inv[i]=(P-P/i)*Inv[P%i]%P;
	}
	rep(i,2,N-1) Inv[i]=Inv[i-1]*Inv[i]%P;
	Tr[1]=Tr[2]=1;
	rep(i,3,N-1) Tr[i]=qpow(i,i-2); // n个点树的数量
	dp[0]=1;
	Solve(0,n);
	int R=1,cc=-1;
	while(R<=n*2) R<<=1,cc++;
	rep(i,1,R) rev[i]=(rev[i>>1]>>1)|((i&1)<<cc);
	rep(i,0,R) A[i]=B[i]=0;
	rep(i,0,n) A[i]=dp[i]*Inv[i]%P;
	rep(i,2,n) B[i]=Tr[i]*Inv[i-2]%P;
	NTT(R,A,1),NTT(R,B,1);
	rep(i,0,R-1) A[i]=A[i]*B[i]%P;
	NTT(R,A,-1);
	f[1]=1;
	rep(i,2,n) f[i]=A[i]*Fac[i-2]%P; // 计算1，n联通的方案数
	rep(kase,1,rd()) {
		n=rd();
		ll s1=1,s2=1;
		rep(i,1,n) {
			int x=rd();
			s1=s1*dp[x]%P;
			s2=s2*f[x]%P;
		}
		ll ans=(qpow(2,n))*s1-s2;
		ans=(ans%P+P)%P;
		printf("%lld\n",ans);
	}
}