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<bits/stdc++.h> 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<n;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<int> #define pi pair<int,int>#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<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #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;}