有标号DAG计数 [容斥原理 子集反演 组合数学 fft]

有标号DAG计数

题目在COGS上

[HZOI 2015]有标号的DAG计数 I

[HZOI 2015] 有标号的DAG计数 II

[HZOI 2015]有标号的DAG计数 III


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。\(n \le 5000, n \le 10^5\)

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);
}


posted @ 2017-05-03 18:17  Candy?  阅读(1132)  评论(0编辑  收藏  举报