题解 点双连通图计数
题目大意
给出\(n\)，求出\(n\)个点的图满足该图为一个点双连通分量的方案数。
前置知识
-
拓展拉格朗日反演
-
多项式指数函数、对数函数
思路
如果做过有标号无向连通图计数就最好了。
我们来重温一下，我们设有标号无向图的指数生成函数为\(F(x)\)，可以得到:
\[F(x)=\sum_{i=0}^{\infty} \frac{2^{\binom{i}{2}}}{i!}x^i
\]
这里的\(2^{\binom{i}{2}}\)就是每条边要或者不要。
我们再设有标号无向连通图的指数生成函数为\(G(x)\)，可以得到:
\[G(x)=\sum_{i=0}^{\infty} \frac{F^i(x)}{i!}=e^{F(x)}
\]
\[\Rightarrow F(x)=\ln G(x)
\]
我们现在再设\(D(x)\)为有标号有根无向连通图，可以得到:
\[[x^n]D(x)=n[x^n]G(x)
\]
我们再设\(b_i\)为\(i\)个点的无根点双联通分量的个数，我们可以得到：
\[D(x)=xe^{\sum_{i=1}^{\infty} b_{i+1}\frac{D^i(x)}{i!}}
\]
感性理解就是说，一个有标号有根无向连通图我们把根所在的点双提出来，上面每个点对应了一个连通块的根，除以\(i!\)是因为图是不讲顺序的。最重要的一点是每个联通块之间互不干涉，所以不会影响根所在的点双连通分量大小。
如果我们设：
\[B(x)=\sum_{i} b_{i+1} \frac{x^i}{i!}
\]
那我们就可以得到：
\[D(x)=xe^{B(D(x))}
\]
\[\Rightarrow B(D(x))=\ln \frac{D(x)}{x}
\]
\[\Rightarrow B(x)=\ln \frac{x}{D^{-1}(x)}
\]
这里的\(D^{-1}(x)\)是指的\(D(x)\)的复合逆函数，即:
\[D^{-1}(D(x))=x
\]
我们设\(H(x)=\ln \frac{D(x)}{x}\)，可以得到:
\[B(x)=H(D^{-1}(x))
\]
这里使用拓展拉格朗日反演公式可以得到：
\[[x^n]B(x)=[x^n]H(D^{-1}(x))
\]
\[=\frac{1}{n}[x^{-1}]H^{'}(x)(\frac{1}{D(x)})^n
\]
\[=\frac{1}{n}[x^{n-1}]H^{'}(x)(\frac{x}{D(x)})^n
\]
这里使用多项式快速幂公式就可以得到：
\[=\frac{1}{n}[x^{n-1}]H^{'}(x)e^{-n\ln \frac{D(x)}{x}}
\]
\[=\frac{1}{n}[x^{n-1}]H^{'}(x)e^{-nH(x)}
\]
于是，我们就可以在\(\Theta(n\log n)\)的时间复杂度内解决这个问题了。不过不知道为什么我的常熟异常的大，开了\(\text {O}_2\)还没有别人不开\(\text {O}_2\)跑得快。为我自己默哀。。。
\(\text {Code}\)
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define mod 998244353
#define Gii 332748118
#define ll long long
#define MAXN 600005
#define Gi 3
int quick_pow (int a,int b){
int res = 1;for (;b;b >>= 1,a = 1ll * a * a % mod) if (b & 1) res = 1ll * res * a % mod;
return res;
}
int limit,l,r[MAXN];
void NTT (int *a,int type){
for (Int i = 0;i < limit;++ i) if (i < r[i]) swap (a[i],a[r[i]]);
for (Int mid = 1;mid < limit;mid <<= 1){
int Wn = quick_pow (type == 1 ? Gi : Gii,(mod - 1) / (mid << 1));
for (Int R = mid << 1,j = 0;j < limit;j += R){
for (Int k = 0,w = 1;k < mid;++ k,w = 1ll * w * Wn % mod){
int x = a[j + k],y = 1ll * w * a[j + k + mid] % mod;
a[j + k] = (x + y) % mod,a[j + k + mid] = (x + mod - y) % mod;
}
}
}
if (type == 1) return ;
int Inv = quick_pow (limit,mod - 2);
for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * Inv % mod;
}
int c[MAXN];
void Solve (int len,int *a,int *b){
if (len == 1) return b[0] = quick_pow (a[0],mod - 2),void ();
Solve ((len + 1) >> 1,a,b);
limit = 1,l = 0;
while (limit < (len << 1)) limit <<= 1,l ++;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
for (Int i = 0;i < len;++ i) c[i] = a[i];
for (Int i = len;i < limit;++ i) c[i] = 0;
NTT (c,1);NTT (b,1);
for (Int i = 0;i < limit;++ i) b[i] = 1ll * b[i] * (2 + mod - 1ll * c[i] * b[i] % mod) % mod;
NTT (b,-1);
for (Int i = len;i < limit;++ i) b[i] = 0;
}
void deravitive (int *a,int n){
for (Int i = 1;i <= n;++ i) a[i - 1] = 1ll * a[i] * i % mod;
a[n] = 0;
}
void inter (int *a,int n){
for (Int i = n;i >= 1;-- i) a[i] = 1ll * a[i - 1] * quick_pow (i,mod - 2) % mod;
a[0] = 0;
}
int b[MAXN];
void Ln (int *a,int n){
memset (b,0,sizeof (b));
Solve (n,a,b);deravitive (a,n);
while (limit <= n) limit <<= 1,l ++;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
NTT (a,1),NTT (b,1);
for (Int i = 0;i < limit;++ i) a[i] = 1ll * a[i] * b[i] % mod;
NTT (a,-1),inter (a,n);
for (Int i = n + 1;i < limit;++ i) a[i] = 0;
}
int F0[MAXN];
void Exp (int *a,int *B,int n)
{
if (n == 1) return B[0] = 1,void ();
Exp (a,B,(n + 1) >> 1);
for (Int i = 0;i < limit;++ i) F0[i] = B[i];
Ln (F0,n);
F0[0] = (a[0] + 1 + mod - F0[0]) % mod;
for (Int i = 1;i < n;++ i) F0[i] = (a[i] + mod - F0[i]) % mod;
NTT (F0,1);NTT (B,1);
for (Int i = 0;i < limit;++ i) B[i] = 1ll * F0[i] * B[i] % mod;
NTT (B,-1);
for (Int i = n;i < limit;++ i) B[i] = 0;
}
int read ()
{
int x = 0;char c = getchar();int f = 1;
while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}
while (c >= '0' && c <= '9'){x = (x << 3) + (x << 1) + c - '0';c = getchar();}
return x * f;
}
void write (int x)
{
if (x < 0){x = -x;putchar ('-');}
if (x > 9) write (x / 10);
putchar (x % 10 + '0');
}
int fac[MAXN],caf[MAXN],lim = 3e5;
void init (){
fac[0] = 1;for (Int i = 1;i <= lim;++ i) fac[i] = 1ll * fac[i - 1] * i % mod;
caf[lim] = quick_pow (fac[lim],mod - 2);for (Int i = lim;i;-- i) caf[i - 1] = 1ll * caf[i] * i % mod;
}
int H[MAXN],H_[MAXN],G[MAXN],FG[MAXN];
void prepare (){
int len = 1 << 17,lll = 17;
for (Int i = 0;i < len;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << lll - 1);
for (Int i = 0;i < len;++ i) H[i] = 1ll * quick_pow (2,1ll * i * (i - 1) / 2 % (mod - 1)) * caf[i] % mod;
Ln (H,len - 1);
for (Int i = 0;i < len;++ i) H[i] = 1ll * H[i + 1] * (i + 1) % mod;H[len - 1] = 0;
Ln (H,len - 1);
for (Int i = 0;i < len;++ i) H_[i] = H[i];limit = 1 << 18,l = 18;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << l - 1);
deravitive (H_,len - 1),NTT (H_,1);
}
void work (int n){
if (!(-- n)) return puts ("1"),void ();
limit = 1 << 17,l = 17;
memset (F0,0,sizeof (F0)),memset (FG,0,sizeof (FG));
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << l - 1);
for (Int i = 0;i < limit;++ i) G[i] = 1ll * H[i] * (mod - n) % mod;
Exp (G,FG,1 << 17),limit = 1 << 18,l = 18;
for (Int i = 0;i < limit;++ i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << l - 1);
NTT (FG,1);for (Int i = 0;i < limit;++ i) FG[i] = 1ll * FG[i] * H_[i] % mod;NTT (FG,-1);
write (1ll * FG[n - 1] * fac[n - 1] % mod),putchar ('\n');
}
signed main(){
init (),prepare ();
for (Int i = 1;i <= 5;++ i) work (read ());
return 0;
}
