[有向图树形图计数] BZOJ 4894 天赋

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~~根本不知道题面在讲什么~~
矩阵树定理有向图版本邻接矩阵还是邻接矩阵度数矩阵根据是出度还是入度分别计算内向树和外向树

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long long ll;const int N=305;const int P=1e9+7;int n;char s[N][N];inline ll 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;}inline ll Inv(ll a){  return Pow(a,P-2);}int a[N][N];inline int det(int n){  int f=0;  for (int i=1;i<=n;i++){    int k=0;    for (int j=i;j<=n;j++) if (a[j][i]) {k=j; break; }    if (i^k) { for (int j=i;j<=n;j++) swap(a[i][j],a[k][j]); f^=1; }    for (int j=i+1;j<=n;j++){      ll t=(ll)Inv(a[i][i])*a[j][i]%P;      for (int k=i;k<=n;k++)    (a[j][k]+=P-(ll)t*a[i][k]%P)%=P;    }  }  ll ret=1;  for (int i=1;i<=n;i++) ret=ret*a[i][i]%P;  return f?(P-ret)%P:ret;}int main(){  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d",&n);  for (int i=1;i<=n;i++) scanf("%s",s[i]+1);  for (int i=1;i<=n;i++)    for (int j=1;j<=n;j++)      if (s[i][j]=='1'){    a[n-i+1][n-j+1]=P-1;    a[n-j+1][n-j+1]++;      }  int ans=det(n-1);  printf("%d\n",ans);  return 0;}

还有个我以前一直以为只能这么做的状压做法

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