【ZOJ】3874 Permutation Graph 【FFT+CDQ分治】

傳送門：【ZOJ】3874 Permutation Graph

題目分析：

容易知道一個個連通塊内部的标号都是連續的，否則一定會有另一個連通塊向這個連通塊建邊，或者這個連通塊向另一個連通塊建邊。而且從左到右左邊的連通塊内最大的标号小于右邊連通塊内最小的标号。

然後我們可以構造dp方程：

dp[n]=n!−i!∗dp[n−i]

容易發現右邊存在一個卷積，由于給的素數正好是費馬素數，是以可以用 NTT 來求卷積，保證精度。因為我們要求所有的dp值，是以我們要用 CDQ 分治優化。

my code：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )

const int P =  ;
const int MAXN =  ;
const int g =  ;

LL power ( int a , int b , LL res =  ) {
    for ( LL tmp = a ; b ; b >>=  , tmp = tmp * tmp % P ) if ( b &  ) res = res * tmp % P ;
    return res ;
}

void NTT ( LL y[] , int n , int rev ) {
    LL inv = power ( n , P -  ) , wn , w , t ;
    for ( int i =  , j , k , t ; i < n ; ++ i ) {
        for ( j =  , k = n >>  , t = i ; k ; k >>=  , t >>=  ) j = j <<  | t &  ;
        if ( i < j ) swap ( y[i] , y[j] ) ;
    }
    for ( int s =  , ds =  , k , i ; s <= n ; ds = s , s <<=  ) {
        wn = power ( g , ( P -  ) / s ) ;
        if ( rev == - ) wn = power ( wn , P -  ) ;
        for ( k =  ; k < n ; k += s ) {
            for ( i = k , w =  ; i < k + ds ; ++ i , w = w * wn % P ) {
                y[i + ds] = ( y[i] - ( t = w * y[i + ds] % P ) + P ) % P ;
                y[i] = ( y[i] + t ) % P ;
            }
        }
    }
    if ( rev == - ) for ( int i =  ; i < n ; ++ i ) y[i] = y[i] * inv % P ;
}

LL x1[MAXN] , x2[MAXN] ;
LL f[MAXN] , dp[MAXN] ;

void cdq ( int l , int r ) {
    if ( l +  == r ) return ;
    int m = ( l + r ) >>  , n =  ;
    cdq ( l , m ) ;
    while ( n < r - l +  ) n <<=  ;
    for ( int i =  ; i < m - l ; ++ i ) x1[i] = dp[i + l] ;
    for ( int i = m - l ; i < n ; ++ i ) x1[i] =  ;
    for ( int i =  ; i < n ; ++ i ) x2[i] = f[i] ;
    NTT ( x1 , n ,  ) ;
    NTT ( x2 , n ,  ) ;
    for ( int i =  ; i < n ; ++ i ) x1[i] = x1[i] * x2[i] % P ;
    NTT ( x1 , n , - ) ;
    for ( int i = m ; i < r ; ++ i ) dp[i] = ( dp[i] - x1[i - l] + P ) % P ;
    cdq ( m , r ) ;
}

void preprocess () {
    f[] =  ;
    for ( int i =  ; i < MAXN ; ++ i ) {
        f[i] = f[i - ] * ( LL ) i % P ;
        dp[i] = f[i] ;
    }
    cdq ( 1 , 100001 ) ;
    //for ( int i =  ; i <=  ; ++ i ) printf ( "%lld\n" , dp[i] ) ;
}

void solve () {
    int x , c , n ;
    int ans =  ;
    scanf ( "%*d%d" , &n ) ;
    for ( int i =  ; i <= n ; ++ i ) {
        scanf ( "%d" , &x ) ;
        int minv =  , maxv =  ;
        for ( int j =  ; j < x ; ++ j ) {
            scanf ( "%d" , &c ) ;
            minv = min ( c , minv ) ;
            maxv = max ( c , maxv ) ;
        }
        if ( maxv - minv +  != x ) ans =  ;
        ans = ans * dp[x] % P ;
    }
    printf ( "%d\n" , ans ) ;
}

int main () {
    int T ;
    preprocess () ;
    scanf ( "%d" , &T ) ;
    for ( int i =  ; i <= T ; ++ i ) solve () ;
    return  ;
}