HDU5552 Bus Routes (CDQ+NTT)

HDU-5552 Bus Routes (CDQ+NTT)

这道题的本质其实是求\(n\)个点带环联通图的数量

实际上就是\(n\)个点联通图的数量减去树的数量

树的数量可以通过\(Prufer\)序列得到是\(n^{n-2}\)，这个东西去问度娘吧

\(n\)个点联通图的数量你可以去做 BZOJ-3456 详细题解

注意这道题比较特殊，\(n\)个点随意的方案数是\((m+1)^{n(n-1)/2}\)

#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)

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

const ll N=2e4+10,P=152076289;
const ll G=106,IG=101862420;

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;
}

int n,m;
ll Inv[N],Fac[N],Pow[N];
ll dp[N],g[N];
int rev[N];
void NTT(int n,ll *a,int f) {
	rep(i,1,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?G:IG,(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)%P;
				a[j]=(a[j]+t)%P;
			}
		}
	}
	if(f==-1) {
		ll base=qpow(n,P-2);
		rep(i,0,n-1) a[i]=a[i]*base%P;
	}
}

ll A[N],B[N],C[N];


void Solve(int l,int r){ 
	if(l==r) return;
	int mid=(l+r)>>1;
	Solve(l,mid);
	int R=1,cc=-1;
	while(R<=r-l+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,l,mid) A[i-l]=-dp[i]*Inv[i-1]%P;
	rep(i,1,r-l+1) B[i]=g[i]*Inv[i]%P;
	NTT(R,A,1),NTT(R,B,1);
	rep(i,0,R) 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(){
	Inv[0]=Inv[1]=Fac[0]=Fac[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;
	rep(kase,1,rd()) {
		n=rd(),m=rd();
		rep(i,1,n) g[i]=dp[i]=qpow(m%P+1,i*(i-1)/2);
		Solve(1,n);//CDQ求出联通图的方案数
		ll ans=dp[n]-qpow(m,n-1)*qpow(n,n-2)%P;
		ans=(ans%P+P)%P;
		printf("Case #%d: %lld\n",kase,ans);
	}
}