HDU 6064 RXD and numbers

传送门

有向图生成树计数 （度数 ->入度->外向树）

BEST定理 （不定起点的欧拉回路个数=某点为根的外向树个数（存在欧拉回路->每个点为根的外向树个数相等）*（每个点的度数（存在欧拉回路->每个点入度=出度）-1）的阶层）

一个题解的传送门

 //Achen
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 #include<cstdio>
 #include<queue>
 #include<cmath>
 #include<set>
 #include<map>
 #define Formylove return 0
 #define For(i,a,b) for(int i=(a);i<=(b);i++)
 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
 const int N=,p=;
 typedef long long LL;
 typedef double db;
 using namespace std;
 LL n,d[N][N],fac[],in[N],out[N];

 template<typename T>void read(T &x)  {
     char ch=getchar(); x=; T f=;
     while(ch!='-'&&(ch<''||ch>'')) ch=getchar();
     if(ch=='-') f=-,ch=getchar();
     for(;ch>=''&&ch<='';ch=getchar()) x=x*+ch-''; x*=f;
 }

 LL a[N][N];
 LL gauss(int n) {
     For(i,,n) For(j,,n) (a[i][j]+=p)%=p;
     LL rs=,f=;
     For(i,,n) {
         For(j,i+,n) {
             LL A=a[i][i],B=a[j][i];
             while(B) {
                 LL t=A/B; A%=B; swap(A,B);
                 For(k,i,n) a[i][k]=(a[i][k]-t*a[j][k]%p+p)%p;
                 For(k,i,n) swap(a[i][k],a[j][k]); f=-f;
             }
         }
         rs=rs*a[i][i]%p;
     }
     if(f==-) rs=(p-rs)%p;
     return rs;
 }

 LL ksm(LL a,LL b) {
     LL rs=,bs=a%p;
     while(b) {
         if(b&) rs=rs*bs%p;
         bs=bs*bs%p;
         b>>=;
     }
     return rs;
 } 

 int main() {
 #ifdef ANS
     freopen(".in","r",stdin);
     freopen(".out","w",stdout);
 #endif
     fac[]=; int cas=;
     For(i,,) fac[i]=fac[i-]*i%p;
     while(~scanf("%lld",&n)) {
         cas++;
         For(i,,n) in[i]=out[i]=;
         For(i,,n) For(j,,n) a[i][j]=;
         For(i,,n) For(j,,n) {
             read(d[i][j]);
             a[i][j]-=d[i][j];
             a[j][j]+=d[i][j];
             (out[i]+=d[i][j])%=p;
             (in[j]+=d[i][j])%=p;
         }
         int fl=;
         For(i,,n) if(in[i]!=out[i]) {
             fl=; break;
         }
         if(fl) {
             printf("Case #%d: 0\n", cas);
             continue;
         }
         For(i,,n) For(j,,n) a[i-][j-]=a[i][j];
         LL ans=gauss(n-);
         For(i,,n)
             ans=ans*fac[in[i]-]%p;
         ans=ans*fac[in[]]%p;
         For(i,,n) For(j,,n) if(d[i][j])
             ans=ans*ksm(fac[d[i][j]],p-)%p;
         printf("Case #%d: %lld\n",cas,ans);
     }
     Formylove;
 }
 /*
 5
 0 1 0 0 0
 0 0 1 0 4
 0 0 0 5 0
 1 5 0 0 0
 0 0 0 1 0
 */