有标号DAG计数

题目在COGS上

I

求n个点的DAG(可以不连通)的个数。$n \le 5000$

2013年王迪的论文很详细了

感觉想法很神，自己怎么想到啊？

首先要注意到DAG中一类特殊的点：入度为0的点。以这些点来分类统计

先是一种$O(N^3)$的dp， $d(i,j)$ i个点j个入度为0，转移枚举去掉j个后入度为0点的个数，乘上连边情况

在弱化条件时特殊化

$f(n, S)$ n个点，只有S中的点入度为0

$g(n,S)$ n个点，至少S中的点入度为0

\[g(n, S) = 2^{\mid s\mid (n-\mid S\mid)} g(n-\mid S\mid, \varnothing) \\
g(n, S) = \sum_{S \subset T} f(n, T)\quad (1)
\]

子集反演！

\[f(n, S) = \sum_{S \subset T} (-1)^{\mid T\mid - \mid S\mid} g(n, S)\quad (2)
\]

我们目标是求$g(n, \varnothing)$,

代入$(1),(2)$，然后使用枚举集合大小，乘上组合数的技巧，$m = \mid T\mid, k = \mid S\mid$。还需要用$\binom{n}{k} \binom{k}{m} = \binom{n}{m} \binom{n-m} {k-m}$替换，最后得到

\[g(n, \varnothing) = \sum_{k=1}^n (-1)^{k-1} \binom{n}{k} 2^{k(n-k)} g(n-k, \varnothing)
\]

完成！

现在尝试直接考虑这个式子的意义：

\[n个点DAG个数= \ge 1 个入度为0 - \ge 2个入度为0 + \ge 3....
\]

II

求n个点的DAG(可以不连通)的个数。$n \le 10^5$

当然要用fft啦！分治或者多项式求逆。

$2^{k(n-k)}$怎么办？

和hdu那道题类似，把$(n-k)^2 = n^2 - 2nk + k^2$代入

但这样会出现$2^{\frac{n}{2}}$， 2的逆元$\mod P-1$不存在，所以要求2的二次剩余$x^2 \equiv 2 \pmod {P-1}$

III & IIII

求n个点的DAG(必须连通)的个数。$n \le 5000, n \le 10^5$

和城市规划类似的思想...

可以不连通到连通

Code

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

typedef long long ll;

const int N = 5005, P = 10007;

inline int read() {

    char c=getchar();int x=0,f=1;

    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}

    return x*f;

}

int n, g[N];

int inv[N], fac[N], facInv[N], mi[P+5];

inline int C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}

int main() {

	freopen("DAG.in", "r", stdin);

	freopen("DAG.out", "w", stdout);

	//freopen("in", "r", stdin);

	n = read();

	inv[1] = fac[0] = facInv[0] = 1;

	for(int i=1; i<=n; i++) {

		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;

		fac[i] = fac[i-1] * i %P;

		facInv[i] = facInv[i-1] * inv[i] %P;

	}

	mi[0] = 1;

	for(int i=1; i<=P; i++) mi[i] = mi[i-1] * 2 %P;

	g[0] = 1;

	for(int i=1; i<=n; i++)

		for(int k=1; k<=i; k++) g[i] = (g[i] + ((k&1) ? 1 : -1) * C(i, k) %P * mi[k * (i-k) % (P-1)] %P * g[i-k] %P) %P;

	printf("%d\n", (g[n] + P) %P);

}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

typedef long long ll;

const int N = (1<<18) + 5, P = 998244353, qr2 = 116195171;

inline int read() {

    char c=getchar();int x=0,f=1;

    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}

    return x*f;

}

ll Pow(ll a, int b) {

	ll ans = 1;

	for(; b; b >>= 1, a = a * a %P)

		if(b & 1) ans = ans * a %P;

	return ans;

}

namespace fft {

	int rev[N];

	void dft(int *a, int n, int flag) {

		int k = 0; while((1<<k) < n) k++;

		for(int i=0; i<n; i++) {

			rev[i] = (rev[i>>1]>>1) | ((i&1)<<(k-1));

			if(i < rev[i]) swap(a[i], a[rev[i]]);

		}

		for(int l=2; l<=n; l<<=1) {

			int m = l>>1;

			ll wn = Pow(3, flag == 1 ? (P-1)/l : P-1-(P-1)/l);

			for(int *p = a; p != a+n; p += l)

				for(int k=0, w=1; k<m; k++, w = w*wn%P) {

					int t = (ll) w * p[k+m] %P;

					p[k+m] = (p[k] - t + P) %P;

					p[k] = (p[k] + t) %P;

				}

		}

		if(flag == -1) {

			ll inv = Pow(n, P-2);

			for(int i=0; i<n; i++) a[i] = a[i] * inv %P;

		}

	}

	int t[N];

	void inverse(int *a, int *b, int l) {

		if(l == 1) {b[0] = Pow(a[0], P-2); return;}

		inverse(a, b, l>>1);

		int n = l<<1;

		for(int i=0; i<l; i++) t[i] = a[i], t[i+l] = 0;

		dft(t, n, 1); dft(b, n, 1);

		for(int i=0; i<n; i++) b[i] = (ll) b[i] * (2 - (ll) t[i] * b[i] %P + P) %P;

		dft(b, n, -1); for(int i=l; i<n; i++) b[i] = 0;

	}

}

int n, a[N], b[N], len;

ll inv[N], fac[N], facInv[N];

int main() {

	//freopen("in", "r", stdin);

	freopen("dag_count.in", "r", stdin);

	freopen("dag_count.out", "w", stdout);

	n = read();

	inv[1] = fac[0] = facInv[0] = 1;

	for(int i=1; i<=n; i++) {

		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;

		fac[i] = fac[i-1] * i %P;

		facInv[i] = facInv[i-1] * inv[i] %P;

	}

	for(int i=1; i<=n; i++) {

		int t = facInv[i] * Pow(Pow(qr2, (ll) i * i %(P-1)), P-2) %P;

		if(i&1) b[i] = P - t; else b[i] = t;

	}

	b[0] = (b[0] + 1) %P;

	len = 1; while(len <= n) len <<= 1;

	fft::inverse(b, a, len);

	int ans = (ll) a[n] * fac[n] %P * Pow(qr2, (ll) n * n %(P-1)) %P;

	printf("%d\n", ans);

}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

typedef long long ll;

const int N = 5005, P = 10007;

inline int read() {

    char c=getchar();int x=0,f=1;

    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}

    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}

    return x*f;

}

int n, g[N], f[N];

int inv[N], fac[N], facInv[N], mi[P+5];

inline int C(int n, int m) {return fac[n] * facInv[m] %P * facInv[n-m] %P;}

int main() {

	freopen("DAGIII.in", "r", stdin);

	freopen("DAGIII.out", "w", stdout);

	//freopen("in", "r", stdin);

	n = read();

	inv[1] = fac[0] = facInv[0] = 1;

	for(int i=1; i<=n; i++) {

		if(i != 1) inv[i] = (P - P/i) * inv[P%i] %P;

		fac[i] = fac[i-1] * i %P;

		facInv[i] = facInv[i-1] * inv[i] %P;

	}

	mi[0] = 1;

	for(int i=1; i<=P; i++) mi[i] = mi[i-1] * 2 %P;

	g[0] = 1;

	for(int i=1; i<=n; i++)

		for(int k=1; k<=i; k++)

			g[i] = (g[i] + ((k&1) ? 1 : -1) * C(i, k) %P * mi[k * (i-k) % (P-1)] %P * g[i-k] %P) %P;

	f[0] = 1;

	for(int i=1; i<=n; i++) {

		f[i] = g[i];

		for(int j=1; j<i; j++) f[i] = (f[i] - (ll) C(i-1, j-1) * f[j] %P * g[i-j]) %P;

		if(f[i] < 0) f[i] += P;

	}

	printf("%d\n", (f[n] + P) %P);

}