Codeforces 757C Felicity is coming!

Obviously we can find the answer is the size of each set that can reach each other and then multiply it by the same order. Consider how to find a collection of each complementary phase, that is, a hash. ( However, I have taken 3 modules before )

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <vector>5#include <cstdlib>6#include <cmath>7#include <cstring>8 using namespacestd;9 #defineMAXN 1000010Ten #defineMD (LLG) (1000100000001007) One #defineMD 1000000007 A #defineMD2 (LLG) (5456145545456) - #defineMD3 (LLG) (100007) - #defineLLG Long Long the #defineYyj (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); - LLG n,m,a[maxn],p[maxn],l,r; -  - structnode + { - LLG a1,a2,a3; + }C[MAXN]; A  at BOOLcmpConstNode&a,Constnode&b) - { -     if(a.a1==b.a1 && a.a2==b.a2)returna.a3<b.a3; -     if(A.A1==B.A1)returna.a2<b.a2; -     returna.a1<b.a1; - } in  - intMain () to { +Srand1346); - //yyj ("C"); thep[0]=p[1]=1; *      for(Llg i=2; i<=1000000; i++) p[i]=p[i-1]*i,p[i]%=MD; $Cin>>n>>m;Panax NotoginsengLLG inc=1; -      for(Llg k=1; k<=n;k++) the     { + LLG x; AInc=rand ()%md+1; Llg Inc2=rand ()%md+1, Inc3=rand ()%md+1; thescanf"%i64d",&x); +          for(Llg i=1; i<=x;i++)  -         { $scanf"%i64d",&a[i]); $C[a[i]].a1+=inc; C[A[I]].A2+=INC2; c[a[i]].a3+=inc3; -C[A[I]].A1%=MD; C[A[I]].A2%=MD2; c[a[i]].a3%=md3; -         } theinc*=2; -inc%=MD;Wuyi     } theLLG ans=1; -Sort (c+1, c+m+1, CMP); WuL=1; -      while(l<=m) About     { $R=l; -          while(c[r+1].A1==C[L].A1 && c[r+1].A2==C[L].A2 && c[r+1].A3==C[L].A3 && r+1&LT;=M) r++; -ans*=p[r-l+1]; -ans%=MD; Al=r+1; +     } the //For (Llg i=1;i<=c[0];i++) ans*=n,ans%=md; -cout<<ans; $     return 0; the}

Codeforces 757C Felicity is coming!