6512. 【GDOI2020模拟3.18】树与路径

题目描述

题解

我太难了

见到树+dp+1e5直接刚dp启发式合并

2h中经历了nlogn->n^2->n^3

思想江化

---

考虑把每个点相连的边配对，每配一次就代表把这两段拼起来

n条边配m对的方案为C(n,2)*C(n-2,2)*...*C(n-2(m-1),2)/m!

按哈夫曼树（合并果子）顺序合并多项式，时间O(n log^2 n)

证明（大概）：

每次找长度最小的两个合并，时间为两个的长度（不考虑NTT的log）

设当前剩余k段，则两个长度之和不超2n/(k-1)

若超过，则剩余(k-2)段中最小的不超(k-3)n/((k-1)(k-2))

而两端中较大者不小于n/(k-1)，因此可以用剩余最小的把两段中较大者换掉

那么k取值2~n时，长度和约为n ln n，大概就是log级别

code

#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define ll long long
#define mod 998244353
#define Mod 998244351
#define G 3
#define file
using namespace std;

struct type{
	int x,y;
	friend bool operator < (type a,type b) {return a.y>b.y;}
};
ll a[131072],b[131072],A[131072],jc[100001],Jc[100001],w[100001],s;
int d[100001],n,i,j,k,l,N,len,x,y,n1,n2;
vector<int> f[100001];
priority_queue<type> heap;

ll C(int n,int m)
{
	return jc[n]*Jc[m]%mod*Jc[n-m]%mod;
}

ll qpower(ll a,int b)
{
	ll ans=1;
	
	while (b)
	{
		if (b&1)
		ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	
	return ans;
}

void dft(ll *a,int type)
{
	int i,j,k,l,s1=2,s2=1,S=N;
	ll w,W,u,v;
	
	fo(i,0,N-1)
	{
		j=i;k=0;
		fo(l,1,len)
		k=k*2+(j&1),j>>=1;
		
		A[k]=a[i];
	}
	memcpy(a,A,8*N);
	
	fo(i,1,len)
	{
		if (type==1)
		w=qpower(G,(mod-1)/s1);
		else
		w=qpower(G,(mod-1)-(mod-1)/s1);
		S>>=1;
		
		fo(j,0,S-1)
		{
			W=1;
			
			fo(k,0,s2-1)
			{
				u=a[j*s1+k];
				v=a[j*s1+k+s2]*W;
				
				a[j*s1+k]=(u+v)%mod;
				a[j*s1+k+s2]=(u-v)%mod;
				W=W*w%mod;
			}
		}
		
		s1<<=1;s2<<=1;
	}
}

int main()
{
	freopen("path.in","r",stdin);
	#ifdef file
	freopen("path.out","w",stdout);
	#endif
	
	scanf("%d",&n);
	w[1]=jc[0]=jc[1]=Jc[0]=Jc[1]=1;
	fo(i,2,n)
	{
		w[i]=mod-w[mod%i]*(mod/i)%mod;
		
		jc[i]=jc[i-1]*i%mod;
		Jc[i]=Jc[i-1]*w[i]%mod;
		
		scanf("%d%d",&j,&k);
		++d[j];++d[k];
	}
	fo(i,1,n)
	{
		heap.push({i,d[i]/2});
		
		f[i].push_back(1);
		s=1;
		fo(j,1,d[i]/2)
		{
			s=s*C(d[i]-(j-1)*2,2)%mod;
			f[i].push_back(s*Jc[j]%mod);
		}
		d[i]/=2;
	}
	
	fo(i,1,n-1)
	{
		x=(heap.top()).x;heap.pop();
		y=(heap.top()).x;heap.pop();
		
		n1=d[x]; fo(j,0,n1) a[j]=f[x][j]; f[x].clear();
		n2=d[y]; fo(j,0,n2) b[j]=f[y][j]; f[y].clear();
		
		len=ceil(log2(n1+n2+1));N=qpower(2,len);
		fo(j,n1+1,N-1) a[j]=0;
		fo(j,n2+1,N-1) b[j]=0;
		
		dft(a,1);
		dft(b,1);
		fo(j,0,N-1) a[j]=a[j]*b[j]%mod;
		dft(a,-1);
		
		N=qpower(N,Mod);
		d[x]+=d[y];
		fo(j,0,d[x])
		f[x].push_back(a[j]*N%mod);
		
		heap.push({x,d[x]});
	}
	
	x=(heap.top()).x;
	fo(i,1,n-1)
	if (i<(n-1)-d[x])
	printf("0 ");
	else
	printf("%d ",(f[x][(n-1)-i]+mod)%mod);
	printf("
");
	
	fclose(stdin);
	fclose(stdout);
	
	return 0;
}