写这个套路题只是为了练手 / 借机会整改一下大多项式模板。
直接考虑点分治。
对于当前的一棵子树,考虑求出 \(f_i\) 表示这棵子树的根往下延伸的所有路径中,费用等于 \(i\) 的方案数。
考虑怎么求这个。首先 DFS 求出 \(g_i\) 表示这棵子树的根往下延伸的所有路径中,长度为 \(i\) 的条数。
那么有 \[
f_i = \sum\limits_{j=i}^{n-1} \binom ji g_j
\]
其中 \(n\) 为次数界。
稍微推一下可得 \[
i!f_i = \sum\limits_{j=0}^{n-i-1} \frac1{j!} \cdot (i+j)! g_{i+j}
\]
容易构造卷积形式计算。
求出 \(f_i\) 后,将子树按最大深度排序逐个执行 NTT 合并。
总复杂度 \(O(n \log^2 n)\)。
代码: 1
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using namespace std;
const int N = 1e5;
const int mod = 998244353;
inline int fpow(int a,int b)
{
int ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
return ret;
}
int n;
int to[(N << 1) + 5],pre[(N << 1) + 5],first[N + 5];
inline void add_edge(int u,int v)
{
static int tot = 0;
to[++tot] = v,pre[tot] = first[u],first[u] = tot;
}
namespace Poly
{
const int N = 1 << 18;
const int G = 3;
int lg2[N + 5];
int rev[N + 5],fac[N + 5],ifac[N + 5],inv[N + 5];
int rt[N + 5],irt[N + 5];
inline void init()
{
for(register int i = 2;i <= N;++i)
lg2[i] = lg2[i >> 1] + 1;
int w = fpow(G,(mod - 1) / N);
rt[N >> 1] = 1;
for(register int i = (N >> 1) + 1;i <= N;++i)
rt[i] = (long long)rt[i - 1] * w % mod;
for(register int i = (N >> 1) - 1;i;--i)
rt[i] = rt[i << 1];
fac[0] = 1;
for(register int i = 1;i <= N;++i)
fac[i] = (long long)fac[i - 1] * i % mod;
ifac[N] = fpow(fac[N],mod - 2);
for(register int i = N;i;--i)
ifac[i - 1] = (long long)ifac[i] * i % mod;
for(register int i = 1;i <= N;++i)
inv[i] = (long long)ifac[i] * fac[i - 1] % mod;
}
struct poly
{
vector<int> a;
inline poly(int x = 0)
{
x && (a.push_back(x),1);
}
inline poly(const vector<int> &o)
{
a = o;
shrink();
}
inline void shrink()
{
for(;!a.empty() && !a.back();a.pop_back());
}
inline int size() const
{
return a.size();
}
inline void resize(int x)
{
a.resize(x);
}
inline int operator[](int x) const
{
if(x < 0 || x >= size())
return 0;
return a[x];
}
inline int &operator[](int x)
{
return a[x];
}
inline void clear()
{
vector<int>().swap(a);
}
inline void ntt(int type = 1)
{
int n = size();
type == -1 && (reverse(a.begin() + 1,a.end()),1);
int lg = lg2[n] - 1;
for(register int i = 0;i < n;++i)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg),
i < rev[i] && (swap(a[i],a[rev[i]]),1);
for(register int w = 2,m = 1;w <= n;w <<= 1,m <<= 1)
for(register int i = 0;i < n;i += w)
for(register int j = 0;j < m;++j)
{
int t = (long long)rt[m | j] * a[i | j | m] % mod;
a[i | j | m] = dec(a[i | j],t),a[i | j] = add(a[i | j],t);
}
if(type == -1)
for(register int i = 0;i < n;++i)
a[i] = (long long)a[i] * inv[n] % mod;
}
friend inline poly operator+(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = add(a[i],b[i]);
return poly(ret);
}
friend inline poly operator-(const poly &a,const poly &b)
{
vector<int> ret(max(a.size(),b.size()));
for(register int i = 0;i < ret.size();++i)
ret[i] = dec(a[i],b[i]);
return poly(ret);
}
friend inline poly operator*(poly a,poly b)
{
if(a.a.empty() || b.a.empty())
return poly();
int lim = 1,tot = a.size() + b.size() - 1;
for(;lim < tot;lim <<= 1);
a.resize(lim),b.resize(lim);
a.ntt(),b.ntt();
for(register int i = 0;i < lim;++i)
a[i] = (long long)a[i] * b[i] % mod;
a.ntt(-1),a.shrink();
return a;
}
poly &operator+=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = add(a[i],o[i]);
return *this;
}
poly &operator-=(const poly &o)
{
resize(max(size(),o.size()));
for(register int i = 0;i < o.size();++i)
a[i] = dec(a[i],o[i]);
return *this;
}
poly &operator*=(poly o)
{
return (*this) = (*this) * o;
}
poly deriv() const
{
if(a.empty())
return poly();
vector<int> ret(size() - 1);
for(register int i = 0;i < size() - 1;++i)
ret[i] = (long long)(i + 1) * a[i + 1] % mod;
return poly(ret);
}
poly integ() const
{
if(a.empty())
return poly();
vector<int> ret(size() + 1);
for(register int i = 0;i < size();++i)
ret[i + 1] = (long long)a[i] * inv[i + 1] % mod;
return poly(ret);
}
inline poly modxn(int n) const
{
n = min(n,size());
return poly(vector<int>(a.begin(),a.begin() + n));
}
inline poly inver(int m) const
{
poly ret(fpow(a[0],mod - 2));
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (2 - modxn(k) * ret)).modxn(k);
return ret.modxn(m);
}
inline poly log(int m) const
{
return (deriv() * inver(m)).integ(),modxn(m);
}
inline poly exp(int m) const
{
poly ret(1);
for(register int k = 1;k < m;)
k <<= 1,ret = (ret * (1 - ret.log(k) + modxn(k))).modxn(k);
return ret.modxn(m);
}
};
}
using Poly::init;
using Poly::poly;
int vis[N + 5],sum,sz[N + 5],max_part[N + 5],rt;
int dep[N + 5],mxdep[N + 5];
poly f,g,h,ans;
void get_rt(int p,int fa)
{
sz[p] = 1,max_part[p] = 0;
for(register int i = first[p];i;i = pre[i])
if((to[i] ^ fa) && !vis[to[i]])
get_rt(to[i],p),
sz[p] += sz[to[i]],
max_part[p] = max(max_part[p],sz[to[i]]);
max_part[p] = max(max_part[p],sum - sz[p]);
max_part[p] < max_part[rt] && (rt = p);
}
void get_dep(int p,int fa)
{
mxdep[p] = dep[p];
for(register int i = first[p];i;i = pre[i])
if((to[i] ^ fa) && !vis[to[i]])
dep[to[i]] = dep[p] + 1,get_dep(to[i],p),mxdep[p] = max(mxdep[p],mxdep[to[i]]);
}
void insert(int p,int fa)
{
++h[dep[p]];
for(register int i = first[p];i;i = pre[i])
if((to[i] ^ fa) && !vis[to[i]])
insert(to[i],p);
}
void solve(int p)
{
vis[p] = 1,dep[p] = 0;
vector<int> son;
for(register int i = first[p];i;i = pre[i])
if(!vis[to[i]])
dep[to[i]] = 1,get_dep(to[i],p),son.push_back(to[i]);
sort(son.begin(),son.end(),[](int x,int y)
{
return mxdep[x] < mxdep[y];
});
f.resize(1),f[0] = 1;
for(int v : son)
{
g.resize(mxdep[v] + 1),h.resize(mxdep[v] + 1);
for(register int i = 0;i <= mxdep[v];++i)
h[i] = 0;
insert(v,p);
for(register int i = 0;i <= mxdep[v];++i)
g[i] = (long long)Poly::fac[mxdep[v] - i] * h[mxdep[v] - i] % mod;
for(register int i = 0;i <= mxdep[v];++i)
h[i] = Poly::ifac[i];
h = (g * h).modxn(mxdep[v] + 1);
for(register int i = 0;i <= mxdep[v];++i)
g[i] = (long long)Poly::ifac[i] * h[mxdep[v] - i] % mod;
h = f * g;
h.resize(h.size() + 1);
for(register int i = h.size() - 1;i;--i)
h[i] = add(h[i],h[i - 1]);
ans += h,f += g;
}
g.resize(2),g[0] = g[1] = 1,ans += g;
for(register int i = first[p];i;i = pre[i])
if(!vis[to[i]])
rt = 0,sum = sz[to[i]],get_rt(to[i],p),solve(rt);
}
int main()
{
Poly::init(),max_part[0] = 0x3f3f3f3f;
scanf("%d",&n);
int u,v;
for(register int i = 2;i <= n;++i)
scanf("%d%d",&u,&v),add_edge(u,v),add_edge(v,u);
sum = n,get_rt(1,0),solve(rt);
ans.resize(n + 1);
for(register int i = 0;i <= n;++i)
printf("%d%c",ans[i]," \n"[i == n]);
}