题解 射命丸文的笔记
题目大意
-
定义-竞赛图：任意两点之间有一条有向边的图。
-
定义-哈密顿回路：指除起点和终点外经过所有顶点恰好一次且起点和终点相同的路径。
-
求出\(m\)，求出对于\(\forall n\in[1,m]\)，存在哈密顿回路的竞赛图其中哈密顿回路的期望个数。
-
\(m\le 10^5\)
思路
做这道题主要是练习一下分治\(\texttt{FFT}\)，但是发现了多项式求逆的巨大的时间优势。
我们发现，其实哈密顿回路的总个数其实是很好求的，答案为\((n-1)!2^{\binom{n}{2}-n}\)，解释就是环的个数为\((n-1)!\)个，除了环的边你爱怎么选怎么选。
于是我们的问题就是如何求出至少存在一个哈密顿回路的竞赛图有多少个。我们设\(n\)个点的至少存在一个哈密顿回路的竞赛图有多少个，我们可以得到转移式:
\[f[n]=2^{\binom{n}{2}}-\sum_{i=1}^{n-1} f[i]2^{\binom{n-i}{2}}\binom{n}{i}
\]
解释就是简单容斥，我们先求出总的竞赛图个数，然后减去不存在哈密顿回路的个数。我们枚举\(i\)个点为一个哈密顿回路，然后其他点往这\(i\)个点往连边，然后其他点之间随便怎么连都可以。
这个时候有两种方法可以做。第一种比较显然，就是直接分治\(\texttt {FFT}\)，时间复杂度\(\Theta(n\log^2 n)\)，代码当然也非常好写。
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline", "no-stack-protector", "unroll-loops")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 300005
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
typedef vector <int> poly;
int w[MAXN],rev[MAXN];
void init_ntt (){
int lim = 1 << 18;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
static unsigned long long d[MAXN];
for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
for (Int i = 1;i < lim;i <<= 1)
for (Int j = 0;j < lim;j += i << 1)
for (Int k = 0;k < i;++ k){
int x = d[i + j + k] * w[i + k] % mod;
d[i + j + k] = d[j + k] + mod - x,d[j + k] += x;
}
for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
if (type == -1){
reverse (a.begin() + 1,a.begin() + lim);
for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
}
#undef G
#undef Gi
}
poly operator * (poly a,poly b){
int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
a.resize (lim),b.resize (lim);
ntt (a,lim,1),ntt (b,lim,1);
for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
ntt (a,lim,-1),a.resize (d);
return a;
}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
poly F,G,f,g;
int n,fac[MAXN],ifac[MAXN];
void cdq (int l,int r){
if (l == r)return F[l] = dec (G[l],F[l]),void ();
int mid = (l + r) >> 1;cdq (l,mid);
f.resize (mid - l + 1);for (Int i = l;i <= mid;++ i) f[i - l] = F[i];
g.resize (r - l + 1);for (Int i = 1;i <= r - l;++ i) g[i] = G[i];
f = f * g;for (Int i = mid + 1;i <= r;++ i) F[i] = add (F[i],f[i - l]);cdq (mid + 1,r);
}
signed main(){
init_ntt(),read (n);
fac[0] = 1;for (Int i = 1;i <= n;++ i) fac[i] = mul (fac[i - 1],i);
ifac[n] = inv (fac[n]);for (Int i = n;i;-- i) ifac[i - 1] = mul (ifac[i],i);
F.resize (n + 1),G.resize (n + 1);for (Int i = 1;i <= n;++ i) G[i] = mul (qkpow (2,1ll * i * (i - 1) / 2 % (mod - 1)),ifac[i]);cdq (0,n);
for (Int i = 1;i <= n;++ i){
F[i] = mul (F[i],fac[i]);
if (i == 1) puts ("1");else if (i == 2) puts ("-1");else write (mul (fac[i - 1],mul (qkpow (2,1ll * i * (i - 3) / 2 % (mod - 1)),inv (F[i])))),putchar ('\n');
}
return 0;
}
另外一种办法便是使用多项式求逆。我们发现递推式可以化成这个样子:
\[\sum_{i=1}^{n} f[i]\binom{n}{i}2^{\binom{n-i}{2}}=2^{\binom{n}{2}}
\]
这样可以主要是因为\(\binom{0}{2}=0\)。
然后我们发现如果我们设:
\[G(x)=\sum_{i=1}^{\infty} 2^{\binom{i}{2}}\frac{x^i}{i!}
\]
那么可以得到:
\[G(x)=G(x)F(x)+[x^0]G(x)
\]
其中\(F(x)\)是\(f[1,2,...,n]\)的\(\texttt{EGF}\)。
于是，我们就可以得到:
\[F(x)=1-\frac{[x^0]G(x)}{G(x)}
\]
时间复杂度\(\Theta(n\log n)\)。代码也很简单（只要你提前封装好了）
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline", "no-stack-protector", "unroll-loops")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 300005
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
typedef vector <int> poly;
int w[MAXN],rev[MAXN];
void init_ntt (){
int lim = 1 << 18;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
static unsigned long long d[MAXN];
for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
for (Int i = 1;i < lim;i <<= 1)
for (Int j = 0;j < lim;j += i << 1)
for (Int k = 0;k < i;++ k){
int x = d[i + j + k] * w[i + k] % mod;
d[i + j + k] = d[j + k] + mod - x,d[j + k] += x;
}
for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
if (type == -1){
reverse (a.begin() + 1,a.begin() + lim);
for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
}
#undef G
#undef Gi
}
poly operator * (poly a,poly b){
int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
a.resize (lim),b.resize (lim);
ntt (a,lim,1),ntt (b,lim,1);
for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
ntt (a,lim,-1),a.resize (d);
return a;
}
poly inv (poly a,int n){
poly b(1,inv (a[0])),c;
for (Int l = 4;(l >> 2) < n;l <<= 1){
c.resize (l >> 1);
for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
c.resize (l),b.resize (l);
ntt (c,l,1),ntt (b,l,1);
for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
ntt (b,l,-1),b.resize (l >> 1);
}
b.resize (n);
return b;
}
poly inv (poly a){return inv (a,SZ (a));}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
poly F;
int n,fac[MAXN],ifac[MAXN];
signed main(){
init_ntt(),read (n);
fac[0] = 1;for (Int i = 1;i <= n;++ i) fac[i] = mul (fac[i - 1],i);
ifac[n] = inv (fac[n]);for (Int i = n;i;-- i) ifac[i - 1] = mul (ifac[i],i);
F.resize (n + 1);for (Int i = 0;i <= n;++ i) F[i] = mul (qkpow (2,1ll * i * (i - 1) / 2 % (mod - 1)),ifac[i]);
F = inv (F);for (Int i = 0;i <= n;++ i) F[i] = mul (mod - F[i],fac[i]);F[0] = add (F[0],1);
for (Int i = 1;i <= n;++ i) if (i == 1) puts ("1");else if (i == 2) puts ("-1");else write (mul (fac[i - 1],mul (qkpow (2,1ll * i * (i - 3) / 2 % (mod - 1)),inv (F[i])))),putchar ('\n');
return 0;
}
两者其实时间差距还是蛮大的，附上一张对比图吧。
分治\(\texttt{FFT}\)：
多项式求逆:
