BZOJ 4894(天赋-有向图生成树计数)

有向图基尔矩阵树定理。
注意删除的一行一列必须是根所在的那行那列

#include 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector 
#define pi pair
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<' ';\
                        cout<#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define MAXN (305)
int n;
char s[MAXN][MAXN];
ll pow2(ll a,ll b){
    ll c=1;
    for(;b;b>>=1,a=a*a%F) {
        if (b&1) c=a*c%F;
    }
    return c;
}
ll a[MAXN][MAXN];
ll inv(ll a){return pow2(a,F-2);}
int det(int n) {
    bool f=0;
    For(i,n) {
        int k=0;
        Fork(j,i,n) if (a[j][i]) {k=j;break;}
        For(j,n) swap(a[i][j],a[k][j]);
        if (i^k) f^=1;
        Fork(j,i+1,n) {
            ll t=a[j][i]*inv(a[i][i])%F; 
            Fork(k,i,n) {
                a[j][k]+=F-t*a[i][k]%F;
                a[j][k]%=F;
            }
        }
    }
    ll r=1;
    For(i,n) r=r*a[i][i]%F;
    if (f) r=F-r;
    return r%F;
}
int main()
{
//  freopen("bzoj4894.in","r",stdin);
//  freopen(".out","w",stdout);
    int n=read();
    MEM(a)
    For(i,n) scanf("%s",s[i]+1);
    For(i,n) For(j,n) if (s[i][j]=='1'){
        a[n-i+1][n-j+1]=F-1;
        a[n-j+1][n-j+1]++;
    }
    printf("%d\n",det(n-1));


    return 0;
}