• 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
    ", 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
    ",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 */
    View Code
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  • 原文地址:https://www.cnblogs.com/Achenchen/p/9499687.html
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