[Explanations] a labeled DAG count 3

[ HZOI, 2015] a labeled DAG count III

We already know \ (f_i \) expressed do not necessarily need Unicom \ (i \) DAG node program, consider merging

Reference [explanations] P4841 urban planning (exponential generating function + polynomial Ln) , then the answer \ (h_i \) generating function \ (H (x) \) so the solution

Because of
\ [H (x) = \
sum_ {i = 0} ^ {\ inf} \ dfrac {(F (x)) ^ i} {i!} \] Is
\ [H (x) = e ^ {F (x)} \]
ball \ (\ LN \) is the IV, it can not for direct manual analog \ (F (x) ^ i / i! \)

//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
 
 
using namespace std;  typedef long long ll;
inline int qr(){
      register int ret=0,f=0;
      register char c=getchar();
      while(c<48||c>57)f|=c==45,c=getchar();
      while(c>=48&&c<=57) ret=ret*10+c-48,c=getchar();
      return f?-ret:ret;
}
 
const int maxn=5e3+5;
const int mod=10007;
int c[maxn][maxn];
int dp[maxn];
int f[maxn];
int bin[maxn*maxn];
 
int main(){
      freopen("DAGIII.in","r",stdin);
      freopen("DAGIII.out","w",stdout);
      int n=qr();
      bin[0]=1;dp[0]=1;
      for(register int t=0;t<=n;++t){
        c[t][0]=1;
        for(register int i=1;i<=t;++i){
          c[t][i]=(c[t-1][i-1]+c[t-1][i])%mod;
        }
      }
      for(register int t=1;t<=n*n;++t) bin[t]=(bin[t-1]<<1)%mod;
      
      for(register int t=1;t<=n;++t){
        for(register int i=1,d;i<=t;++i){
          d=mod-c[t][i]*bin[i*(t-i)]%mod*dp[t-i]%mod;
          if(i&1) d=mod-d;
          dp[t]=(dp[t]+d)%mod;
        }
      }
      for(register int t=1;t<=n;++t){
        int d=0;
        for(register int i=1;i<=t;++i)
          d=(d+c[t-1][i-1]*f[i]%mod*dp[t-i]%mod)%mod;
        f[t]=(dp[t]-d+mod)%mod;
      }
      printf("%d\n",f[n]);
      return 0;
}