ZOJ 3874 Permutation Graph (分治NTT优化DP)

  1 #include <cmath>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #define N1 (1<<18)+10
  6 #define il inline
  7 #define dd double
  8 #define ld long double
  9 #define ll long long
 10 using namespace std;
 11 
 12 const int inf=0x3f3f3f3f;
 13 const ll p=786433;
 14 int gint()
 15 {
 16     int ret=0,fh=1;char c=getchar();
 17     while(c<‘0‘||c>‘9‘){if(c==‘-‘)fh=-1;c=getchar();}
 18     while(c>=‘0‘&&c<=‘9‘){ret=ret*10+c-‘0‘;c=getchar();}
 19     return ret*fh;
 20 }
 21 ll qpow(ll x,ll y)
 22 {
 23     ll ans=1;
 24     for(;y;x=x*x%p,y>>=1) if(y&1) ans=ans*x%p;
 25     return ans;
 26 }
 27 
 28 namespace NTT{
 29 
 30 ll a[N1],b[N1],c[N1],invwn[N1],mulwn[N1];
 31 int r[19][N1];
 32 void Pre(int len,int L)
 33 {
 34     int i,j;
 35     for(j=1;j<=L;j++) for(i=0;i<(1<<j);i++)
 36         r[j][i]=(r[j][i>>1]>>1)|((i&1)<<(j-1));
 37     for(i=2;i<=len;i<<=1) mulwn[i]=qpow(10,(p-1)/i), invwn[i]=qpow(mulwn[i],p-2);
 38 }
 39 void NTT(ll *s,int len,int type,int L)
 40 {
 41     int i,j,k; ll wn,w,t;
 42     for(i=0;i<len;i++) if(i<r[L][i]) swap(s[i],s[r[L][i]]);
 43     for(k=2;k<=len;k<<=1)
 44     {
 45         wn=(type>0)?mulwn[k]:invwn[k];
 46         for(i=0;i<len;i+=k)
 47         {
 48             for(j=0,w=1;j<(k>>1);j++,w=w*wn%p)
 49             {
 50                 t=w*s[i+j+(k>>1)]%p;
 51                 s[i+j+(k>>1)]=(s[i+j]+p-t)%p;
 52                 s[i+j]=(s[i+j]+t)%p;
 53             }
 54         }
 55     }
 56 }
 57 void Main(int len,int L)
 58 {
 59     int i,invl=qpow(len,p-2);
 60     NTT(a,len,1,L); NTT(b,len,1,L);
 61     for(i=0;i<len;i++) c[i]=a[i]*b[i]%p;
 62     NTT(c,len,-1,L); 
 63     for(i=0;i<len;i++) c[i]=c[i]*invl%p;
 64 }
 65 void clr(int sz)
 66 {
 67     memset(a,0,sz<<3);
 68     memset(b,0,sz<<3);
 69 }
 70 
 71 };
 72 
 73 using NTT::a; using NTT::b; using NTT::c;
 74 ll f[N1],g[N1]; int de;
 75 void CDQ(int l,int r)
 76 {
 77     if(r-l<1) return;
 78     if(r-l==1){ f[l]=(g[l]+p-f[l])%p; return; }
 79     int mid=(l+r)>>1,i,len,L;
 80     CDQ(l,mid);
 81     for(len=1,L=0;len<(mid-l)+(r-l)-1;len<<=1,L++);
 82     for(i=l;i<mid;i++) NTT::a[i-l]=f[i];
 83     for(i=0;i<(r-l);i++) NTT::b[i]=g[i];
 84     NTT::Main(len,L);
 85     for(i=mid;i<r;i++) f[i]=(f[i]+NTT::c[i-l])%p;
 86     NTT::clr(len);
 87     CDQ(mid,r);
 88 }
 89 int T,n,m;
 90 int que[N1];
 91 
 92 int main()
 93 {
 94     int i,j,x,y,len,L,mi,ma; ll ans;
 95     scanf("%d",&T); n=100001;
 96     for(i=2,g[1]=1;i<n;i++) g[i]=g[i-1]*i%p;
 97     for(len=1,L=0;len<n+n-1;len<<=1,L++);
 98     NTT::Pre(len,L);
 99     CDQ(0,n);
100     
101     while(T--){
102     
103     n=gint(); m=gint(); ans=1;
104     for(i=1;i<=m;i++)
105     {
106         x=gint(); mi=inf,ma=0;
107         for(j=1;j<=x;j++) que[j]=gint(), mi=min(mi,que[j]), ma=max(ma,que[j]);
108         if(ma-mi+1!=x) ans=0;
109         ans=(ans*f[x])%p;
110     }
111     printf("%lld\n",ans);
112     
113     }
114     return 0;
115 
116 }