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