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 */
