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

  1 //Achen
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<cstring>
  5 #include<cstdlib>
  6 #include<vector>
  7 #include<cstdio>
  8 #include<queue>
  9 #include<cmath>
 10 #include<set>
 11 #include<map>
 12 #define Formylove return 0
 13 #define For(i,a,b) for(int i=(a);i<=(b);i++)
 14 #define Rep(i,a,b) for(int i=(a);i>=(b);i--)
 15 const int N=505,p=998244353;
 16 typedef long long LL;
 17 typedef double db;
 18 using namespace std;
 19 LL n,d[N][N],fac[8000007],in[N],out[N];
 20 
 21 template<typename T>void read(T &x)  {
 22     char ch=getchar(); x=0; T f=1;
 23     while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
 24     if(ch=='-') f=-1,ch=getchar();
 25     for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
 26 }
 27 
 28 LL a[N][N];
 29 LL gauss(int n) {
 30     For(i,1,n) For(j,1,n) (a[i][j]+=p)%=p;
 31     LL rs=1,f=1;
 32     For(i,1,n) {
 33         For(j,i+1,n) {
 34             LL A=a[i][i],B=a[j][i];
 35             while(B) {
 36                 LL t=A/B; A%=B; swap(A,B);
 37                 For(k,i,n) a[i][k]=(a[i][k]-t*a[j][k]%p+p)%p;
 38                 For(k,i,n) swap(a[i][k],a[j][k]); f=-f;
 39             }
 40         }
 41         rs=rs*a[i][i]%p;
 42     }
 43     if(f==-1) rs=(p-rs)%p;
 44     return rs;
 45 }
 46 
 47 LL ksm(LL a,LL b) {
 48     LL rs=1,bs=a%p;
 49     while(b) {
 50         if(b&1) rs=rs*bs%p;
 51         bs=bs*bs%p;
 52         b>>=1;
 53     }
 54     return rs;
 55 } 
 56 
 57 int main() {
 58 #ifdef ANS
 59     freopen(".in","r",stdin);
 60     freopen(".out","w",stdout);
 61 #endif
 62     fac[0]=1; int cas=0;
 63     For(i,1,1000000) fac[i]=fac[i-1]*i%p;
 64     while(~scanf("%lld",&n)) {
 65         cas++;
 66         For(i,1,n) in[i]=out[i]=0;
 67         For(i,1,n) For(j,1,n) a[i][j]=0;
 68         For(i,1,n) For(j,1,n) {
 69             read(d[i][j]);
 70             a[i][j]-=d[i][j];
 71             a[j][j]+=d[i][j];
 72             (out[i]+=d[i][j])%=p;
 73             (in[j]+=d[i][j])%=p;
 74         }
 75         int fl=0;
 76         For(i,1,n) if(in[i]!=out[i]) {
 77             fl=1; break;
 78         }
 79         if(fl) {
 80             printf("Case #%d: 0\n", cas);
 81             continue;
 82         }
 83         For(i,2,n) For(j,2,n) a[i-1][j-1]=a[i][j];
 84         LL ans=gauss(n-1);
 85         For(i,2,n) 
 86             ans=ans*fac[in[i]-1]%p;
 87         ans=ans*fac[in[1]]%p;
 88         For(i,1,n) For(j,1,n) if(d[i][j]) 
 89             ans=ans*ksm(fac[d[i][j]],p-2)%p;
 90         printf("Case #%d: %lld\n",cas,ans);
 91     }
 92     Formylove;
 93 }
 94 /*
 95 5
 96 0 1 0 0 0
 97 0 0 1 0 4
 98 0 0 0 5 0
 99 1 5 0 0 0
100 0 0 0 1 0
101 */