题目
边双联通图:若一个无向图删去任意一条边,这个无向图仍联通,则这个无向图是边双连通图。
分析
同 点双连通图计数,先得到 有根无向连通图 的 EGF ,设 有根边双连通图 的 EGF 为 。相对点双而言,边双比较简单,因为根最多属于一个 极大边双 中(否则,根所在的两个边双显然可以合成一个边双),因此只需考虑根 所在的边双,然后对于一个以 为根的 有根无向连通图,它就是将一些 有根无向连通图 分别通过一条割边挂在了 所在的边双上,因此枚举根所在的 极大边双 的大小,由于每个挂上去的 有根无向连通图 可以选择挂在哪个点,于是可以得到 设 ,则 代入 扩展拉格朗日反演 中可得 直接算即可,注意我们算的是 EGF 并且是有根的,所以最后要除以 乘 。
代码
变量名有出入。
#include <bits/stdc++.h>
#define RG register
typedef long long LL;
int Read() {
int x = 0; bool f = false; char c = getchar();
while (c < '0' || c > '9')
f |= c == '-', c = getchar();
while (c >= '0' && c <= '9')
x = (x * 10) + (c ^ 48), c = getchar();
return f ? -x : x;
}
template <const int _MOD> struct ModNumber { // 为了效率 (事实上还是不高) 省去了一些实用性
int x;
inline ModNumber() { x = 0; }
inline ModNumber(const int &y) { x = y; }
// 需保证 y 的范围! (如果在这 % _MOD 会 T, 因为代码中大量调用该构造函数)
// (当然, 开 O2 可以起飞)
inline int Int() { return x; }
inline ModNumber Pow(LL y) const {
RG int ret = 1, tmp = x;
while (y) {
if (y & 1) ret = ((LL)ret * tmp) % _MOD;
y >>= 1; tmp = ((LL)tmp * tmp) % _MOD;
}
return ModNumber(ret);
}
inline bool operator == (const ModNumber &y) const { return x == y.x; }
inline bool operator != (const ModNumber &y) const { return x != y.x; }
inline bool operator < (const ModNumber &y) const { return x < y.x; }
inline bool operator > (const ModNumber &y) const { return x > y.x; }
inline bool operator <= (const ModNumber &y) const { return x <= y.x; }
inline bool operator >= (const ModNumber &y) const { return x >= y.x; }
inline ModNumber operator + (const ModNumber &y) const { return (x + y.x >= _MOD) ? (x + y.x - _MOD) : (x + y.x); }
inline ModNumber operator - (const ModNumber &y) const { return (x - y.x < 0) ? (x - y.x + _MOD) : (x - y.x); }
inline ModNumber operator * (const ModNumber &y) const { return ModNumber((LL)x * y.x % _MOD); }
inline ModNumber operator / (const ModNumber &y) const { return *this * y.Pow(_MOD - 2); }
inline ModNumber operator ^ (const LL &y) const { return Pow(y); }
inline void operator += (const ModNumber &y) { *this = *this + y; }
inline void operator *= (const ModNumber &y) { *this = *this * y; }
inline void operator -= (const ModNumber &y) { *this = *this - y; }
inline void operator /= (const ModNumber &y) { *this = *this / y; }
inline void operator ^= (const LL &y) const { *this = *this ^ y; }
};
const int MAXN = 100000 * 4; // 所有 MAXN 都要开 4 倍!
const int MOD = 998244353;
typedef ModNumber<MOD> Int;
const Int __G = 3, One = 1, Two = 2, InvTwo = One / Two;
namespace Polynomial {
// 好氧, 好氧, 好氧! (无氧可能原地去世)
// 所有 n: 多项式的项数 (即次数 + 1)
// 数组从 0 开始存
int Rev[MAXN + 5];
Int G0[2][MAXN + 5];
void GetG0(const int &n) { // 使用前必须先初始化 G0
for (RG int i = 2; i <= n; i <<= 1) {
G0[0][i] = __G ^ ((MOD - 1) / i);
G0[1][i] = G0[0][i] ^ (MOD - 2);
}
}
inline void GetRev(const int n) { // Rev 在函数内初始化
for (RG int i = 0; i < n; i++)
Rev[i] = (Rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
}
inline int ToPow(const int &n) { // lim 在函数内初始化
RG int ret = 1;
while (ret < n)
ret <<= 1;
return ret;
}
void PrintPoly(Int *A, const int &n) {
for (RG int i = 0; i < n; i++)
printf("%d ", A[i].x);
puts("");
}
void ReadPoly(Int *A, const int &n) {
for (RG int i = 0; i < n; i++)
A[i].x = Read();
}
void NTT(Int *A, const int &n, const int &opt) {
for (RG int i = 0; i < n; i++)
if (i < Rev[i])
std::swap(A[i], A[Rev[i]]);
for (RG int mid = 1; mid < n; mid <<= 1) {
const int k = mid << 1;
const Int g0 = G0[opt][k];
for (RG int i = 0; i < n; i += k) {
Int g = 1;
for (RG int j = 0; j < mid; j++, g *= g0) {
Int tmp1 = A[i + j], tmp2 = A[i + j + mid] * g;
A[i + j] = tmp1 + tmp2, A[i + j + mid] = tmp1 - tmp2;
}
}
}
if (opt == 1) {
const Int inv = One / n;
for (RG int i = 0; i < n; i++)
A[i] *= inv;
}
}
Int A0[MAXN + 5], B0[MAXN + 5];
void Multiply(const Int *A, const Int *B, Int *P, const int &n, const int &m) { // P = A * B
int lim = ToPow(n + m - 1);
GetRev(lim);
for (int i = 0; i < lim; i++)
A0[i] = B0[i] = 0;
for (RG int i = 0; i < n; i++)
A0[i] = A[i];
for (RG int i = 0; i < m; i++)
B0[i] = B[i];
NTT(A0, lim, 0), NTT(B0, lim, 0);
for (RG int i = 0; i < lim; i++)
P[i] = A0[i] * B0[i];
NTT(P, lim, 1);
}
Int A1[MAXN + 5], B1[MAXN + 5], C1[MAXN + 5];
void Inverse(const Int *A, Int *B, const int &n) { // B = 1 / A, A 不变
if (n == 1) {
B[0] = A[0] ^ (MOD - 2);
return;
}
Inverse(A, B, (n + 1) >> 1);
const int lim = ToPow(n + n - 1);
GetRev(lim);
for (RG int i = 0; i < n; i++)
A1[i] = A[i], B1[i] = B[i];
for (RG int i = n; i < lim; i++)
A1[i] = B1[i] = 0;
NTT(A1, lim, 0), NTT(B1, lim, 0);
for (RG int i = 0; i < lim; i++)
C1[i] = A1[i] * B1[i] * B1[i];
NTT(C1, lim, 1);
for (RG int i = 0; i < n; i++)
B[i] = Two * B[i] - C1[i];
for (RG int i = n; i < lim; i++)
B[i] = 0;
}
Int C2[MAXN + 5], InvB[MAXN + 5], B2[MAXN + 5];
void Sqrt(const Int *A, Int *B, const int &n) { // B = √A, A 不变 (默认开根的多项式常数项为 1)
if (n == 1) {
B[0] = 1;
return;
}
Sqrt(A, B, (n + 1) >> 1);
Inverse(B, InvB, n);
Multiply(B, B, B2, n, n);
for (RG int i = 0; i < n; i++)
InvB[i] *= InvTwo;
for (RG int i = 0; i < n; i++)
B2[i] += A[i];
Multiply(B2, InvB, C2, n, n);
for (RG int i = 0; i < n; i++)
B[i] = C2[i];
}
Int AR[MAXN + 5], BR[MAXN + 5], InvBR[MAXN + 5], A2[MAXN + 5], CR[MAXN + 5], C3[MAXN + 5];
void Divide(const Int *A, const Int *B, Int *C, Int *R, const int &n, const int &m) { // C = A / B, R = A % B, A 不变, B 不变
for (RG int i = 0; i < n; i++)
AR[i] = A[n - i - 1], A2[i] = A[i];
for (RG int i = 0; i < n - m + 1; i++)
BR[i] = B[m - i - 1];
Inverse(BR, InvBR, n - m + 1);
Multiply(AR, InvBR, CR, n, n - m + 1);
for (int i = 0; i < n - m + 1; i++)
C[i] = CR[n - m - i];
Multiply(B, C, C3, m, n - m + 1);
for (RG int i = 0; i < m - 1; i++)
R[i] = A[i] - C3[i];
}
void Derivative(const Int *A, Int *B, const int &n) { // B = A', A 不变
for (RG int i = 0; i < n - 1; i++)
B[i] = A[i + 1] * (i + 1);
B[n - 1] = 0;
}
void Integral(const Int *A, Int *B, const int &n) { // B = ∫A, A 不变
for (RG int i = 1; i < n; i++)
B[i] = A[i - 1] / i;
B[0] = 0;
}
Int AD[MAXN + 5], InvA[MAXN + 5], C4[MAXN + 5];
void Ln(const Int *A, Int *B, const int &n) { // B = ln A, A 不变
Derivative(A, AD, n);
Inverse(A, InvA, n);
Multiply(AD, InvA, C4, n, n);
Integral(C4, B, n);
}
Int LnB[MAXN + 5], B3[MAXN + 5], B4[MAXN + 5];
void Exp(const Int *A, Int *B, const int &n) { // B = e^A, A 不变
if (n == 1) {
B[0] = 1;
return;
}
Exp(A, B, (n + 1) >> 1);
Ln(B, LnB, n);
const int lim = ToPow(n + n - 1);
for (int i = 0; i < n; i++)
B3[i] = A[i] - LnB[i], B4[i] = B[i];
B3[0] += One;
Multiply(B3, B4, B, n, n);
for (int i = n; i < lim; i++)
B[i] = 0;
}
Int KA[MAXN + 5], LnA[MAXN + 5];
void Pow(const Int *A, Int *B, const int &n, const Int &k) { // B = A^k, A 不变
Polynomial::Ln(A, LnA, n);
for (int i = 0; i < n; i++)
KA[i] = LnA[i] * k;
Exp(KA, B, n);
}
Int A3[MAXN + 5], AP[MAXN + 5], C5[MAXN + 5];
Int CompoundInverse(const Int *A, const int &k, const Int *C = NULL) { // A(B(x)) = x 或 A(B(x)) = C(x), 返回 [x^k]B
for (int i = 0; i < k; i++)
A3[i] = A[i + 1];
Pow(A3, AP, k, k);
Inverse(AP, A3, k);
if (C == NULL)
return A3[k - 1] / k;
Derivative(A3, AP, k);
Multiply(AP, C, C5, k, k);
return C5[k - 1] / k;
}
}
int N;
Int Fac[MAXN + 5], Inv[MAXN + 5];
Int G[MAXN + 5], D[MAXN + 5], D1[MAXN + 5], H[MAXN + 5], HE[MAXN + 5], H1[MAXN + 5];
int main() {
Polynomial::GetG0(MAXN);
N = MAXN / 4;
Fac[0] = 1;
for (int i = 1; i <= N; i++)
Fac[i] = Fac[i - 1] * i;
Inv[N] = One / Fac[N];
for (int i = N - 1; i >= 0; i--)
Inv[i] = Inv[i + 1] * (i + 1);
int T = 5;
while (T--) {
if ((N = Read()) == 1) {
puts("1");
continue;
}
G[0] = 1;
for (int i = 1; i <= N; i++)
G[i] = (Two ^ ((LL)i * (i - 1) / 2)) * Inv[i];
Polynomial::Ln(G, D, N + 1);
for (int i = 0; i <= N; i++) {
D[i] = D[i] * i;
H[i] = D[i] * (MOD - N);
}
Polynomial::Derivative(D, D1, N + 1);
Polynomial::Exp(H, HE, N + 1);
Polynomial::Multiply(HE, D1, H1, N + 1, N);
Int Ans = H1[N - 1] / N * Fac[N - 1];
printf("%d\n", Ans.x);
}
return 0;
}