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("\n");
    
    fclose(stdin);
    fclose(stdout);
    
    return 0;
}