[带标号DAG计数 容斥原理 多项式求逆 多项式求ln] COGS 有标号的DAG计数系列

题目链接 传送门I 传送门II 传送门III 传送门IV
题解戳这里
大致解法和思想是同这里一样的

DAG1

找出一个入度为0的点 那么答案为C1n∗21∗(n−1)∗fn−1 但是一张DAG里会有多个入度为0的点会重复计数
考虑容斥 得到

fn=∑k=1n(−1)k−1∗Ckn∗2k∗(n−k)∗fn−k

直接递推求解

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int N=5005;const int P=10007;const int Root2=2641;inline int Pow(int a,int b){  int ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}int _pow[P],C[N][N];inline void Pre(){  C[0][0]=1;  for (int i=1;i<=5000;i++){    C[i][0]=1;    for (int j=1;j<=i;j++)      C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;  }  _pow[0]=1;  for (int i=1;i<P;i++) _pow[i]=_pow[i-1]*2%P;}int n;ll f[N];int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n); Pre();  f[0]=1;  for (int i=1;i<=n;i++){    for (int j=1;j<=i;j++){      int tmp=(ll)C[i][j]*_pow[j*(i-j)%(P-1)]*f[i-j]%P;      if (j&1) f[i]+=tmp; else f[i]-=tmp;    }    f[i]=(f[i]%P+P)%P;  }  printf("%d\n",f[n]);  return 0;}

DAG2

考虑如何拆分2k∗(n−k)

2k∗(n−k)=2n222k22∗2(n−k)22

这里我们需要求2的二次剩余
那么把DAG1中的柿子改为分治FFT或多项式求逆就可以了
但是在DAG1中不能这么做 因为模数不对 汗

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int N=600005;const int P=998244353;const int G=3; const int Root2=116195171;inline int Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}int num;int w[2][N];ll fac[N],inv[N];inline void Pre(int n){  num=n;  ll g=Pow(G,(P-1)/num);  w[1][0]=1; for (int i=1;i<num;i++) w[1][i]=g*w[1][i-1]%P;  w[0][0]=1; for (int i=1;i<num;i++) w[0][i]=w[1][num-i];}int R[N];inline void FFT(int *a,int n,int r){  for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    ll x=a[j+k],y=(ll)w[r][num/(i<<1)*k]*a[j+i+k]%P;    a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;      }  if (!r) for (int i=0,inv=Pow(n,P-2);i<n;i++) a[i]=(ll)a[i]*inv%P;}inline void GetInv(int *a,int *b,int n){  static int tmp[N];  if (n==1) return void(b[0]=Pow(a[0],P-2));  GetInv(a,b,n>>1);  for (int i=0;i<n;i++) tmp[i]=a[i],tmp[n+i]=0;  int L=0; while (!(n>>L&1)) L++;  for (int i=1;i<(n<<1);i++) R[i]=(R[i>>1]>>1)|((i&1)<<L);  FFT(tmp,n<<1,1); FFT(b,n<<1,1);  for (int i=0;i<(n<<1);i++)    tmp[i]=(ll)b[i]*(2+P-(ll)tmp[i]*b[i]%P)%P;  FFT(tmp,n<<1,0);  for (int i=0;i<n;i++) b[i]=tmp[i],b[n+i]=0;}int n,m;int A[N],B[N];int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n);  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;  for (m=1;m<=n;m<<=1); Pre(m<<1);  A[0]=1;  for (int i=1;i<=n;i++){    ll tmp=fac[i]*Pow(Root2,(ll)i*i%(P-1))%P;    tmp=Pow(tmp,P-2);    A[i]=i&1?P-tmp:tmp;  }  GetInv(A,B,m);  int Ans=fac[n]*B[n]%P*Pow(Root2,(ll)n*n%(P-1))%P;  printf("%d\n",Ans);  return 0;}

DAG3

设fn为n个点的DAG的个数
设gn为n个点的连通DAG的个数
容斥的柿子

gn=fn−∑k=1n−1Ck−1n−1∗gk∗fn−k

大力递推

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int N=5005;const int P=10007;const int Root2=2641;inline int Pow(int a,int b){  int ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}int _pow[P],C[N][N];inline void Pre(){  C[0][0]=1;  for (int i=1;i<=5000;i++){    C[i][0]=1;    for (int j=1;j<=i;j++)      C[i][j]=(C[i-1][j-1]+C[i-1][j])%P;  }  _pow[0]=1;  for (int i=1;i<P;i++) _pow[i]=_pow[i-1]*2%P;}int n;ll f[N],g[N];int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n); Pre();  f[0]=1;  for (int i=1;i<=n;i++){    for (int j=1;j<=i;j++){      int tmp=(ll)C[i][j]*_pow[j*(i-j)%(P-1)]*f[i-j]%P;      if (j&1) f[i]+=tmp; else f[i]-=tmp;    }    f[i]=(f[i]%P+P)%P;  }  for (int i=0;i<=n;i++){    g[i]=f[i];    for (int j=1;j<i;j++)      g[i]-=(ll)C[i-1][j-1]*g[j]*f[i-j];    g[i]=(g[i]%P+P)%P;  }  printf("%d\n",g[n]);  return 0;}

DAG4

不论是由容斥的柿子还是多项式exp ln的组合意义都不难得出多项式的做法

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#define cl(x) memset(x,0,sizeof(x))using namespace std;typedef long long ll;const int N=600005;const int P=998244353;const int G=3; const int Root2=116195171;inline int Pow(ll a,int b){  ll ret=1; for (;b;b>>=1,a=a*a%P) if (b&1) ret=ret*a%P; return ret;}int num;int w[2][N];ll fac[N],inv[N];inline void Pre(int n){  num=n;  ll g=Pow(G,(P-1)/num);  w[1][0]=1; for (int i=1;i<num;i++) w[1][i]=g*w[1][i-1]%P;  w[0][0]=1; for (int i=1;i<num;i++) w[0][i]=w[1][num-i];}int R[N];inline void FFT(int *a,int n,int r){  for (int i=0;i<n;i++) if (i<R[i]) swap(a[i],a[R[i]]);  for (int i=1;i<n;i<<=1)    for (int j=0;j<n;j+=(i<<1))      for (int k=0;k<i;k++){    ll x=a[j+k],y=(ll)w[r][num/(i<<1)*k]*a[j+i+k]%P;    a[j+k]=(x+y)%P; a[j+i+k]=(x+P-y)%P;      }  if (!r) for (int i=0,inv=Pow(n,P-2);i<n;i++) a[i]=(ll)a[i]*inv%P;}inline void GetInv(int *a,int *b,int n){  static int tmp[N];  if (n==1) return void(b[0]=Pow(a[0],P-2));  GetInv(a,b,n>>1);  for (int i=0;i<n;i++) tmp[i]=a[i],tmp[n+i]=0;  int L=0; while (!(n>>L&1)) L++;  for (int i=1;i<(n<<1);i++) R[i]=(R[i>>1]>>1)|((i&1)<<L);  FFT(tmp,n<<1,1); FFT(b,n<<1,1);  for (int i=0;i<(n<<1);i++)    tmp[i]=(ll)b[i]*(2+P-(ll)tmp[i]*b[i]%P)%P;  FFT(tmp,n<<1,0);  for (int i=0;i<n;i++) b[i]=tmp[i],b[n+i]=0;}int n,m,F[N];int A[N],B[N],invB[N];inline void Calc(){  A[0]=1;  for (int i=1;i<=n;i++){    ll tmp=fac[i]*Pow(Root2,(ll)i*i%(P-1))%P;    tmp=Pow(tmp,P-2);    A[i]=i&1?P-tmp:tmp;  }  cl(B); GetInv(A,B,m);  for (int i=0;i<=n;i++)    F[i]=fac[i]*B[i]%P*Pow(Root2,(ll)i*i%(P-1))%P;}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n);  fac[0]=1; for (int i=1;i<=n;i++) fac[i]=fac[i-1]*i%P;  inv[1]=1; for (int i=2;i<=n;i++) inv[i]=inv[P%i]*(P-P/i)%P;  inv[0]=1; for (int i=1;i<=n;i++) inv[i]=(inv[i]*inv[i-1])%P;  for (m=1;m<=n;m<<=1); Pre(m<<1);  Calc();  cl(A); cl(B);  for (int i=1;i<=n;i++) A[i]=F[i]*inv[i-1]%P;  for (int i=0;i<=n;i++) B[i]=F[i]*inv[i]%P;  GetInv(B,invB,m);  FFT(A,m<<1,1); FFT(invB,m<<1,1);  for (int i=0;i<(m<<1);i++) A[i]=(ll)A[i]*invB[i]%P;  FFT(A,m<<1,0);  printf("%d\n",(ll)A[n]*fac[n-1]%P);  return 0;}