题目链接
题意简述：
在一个序列中，两点间如果有边，当且仅当两点为逆序对
给定一个序列的联通情况，求方案数对\(786433\)取模
题解
自己弄了一个晚上终于弄出来了
首先\(yy\)一下发现一个很重要的性质：
联通块内的点编号必须是连续的
证明：
假设一个联通块编号不连续，设\(a\)，\(b\)分别为联通块左侧和联通块右侧中的一个点，\(x\)为\(a\)，\(b\)之间不在该联通块内的点
那么显然有\(a > b\)，\(a < x\)，\(x < b\)
即\(a < x < b\)的同时\(a > b\)
不符
故一个联通块内的编号必须连续
证毕
好了我们有了这样一个性质，那么假设他给我们的联通块不符合这个条件，就直接输出\(0\)【一定要记得，我就是一直挂在这个\(sb\)地方调了半天QAQ】
然后如果符合条件，我们就要计算方案数了
因为联通块已经被分成一段一段，所以任意两个联通块之间一定是递增的，互不干涉
所以我们只需要计算出\(f[i]\)表示\(i\)个点联通块的方案
按套路，我们补集转化，并枚举第一个点所在联通块大小
\[f[n] = n! - \sum\limits_{i = 1}^{n - 1} f[i](n - i)!
\]
\(n!\)是总方案，前\(i\)个点联通方案是\(f[i]\)，按照性质，前\(i\)个点一定是前\(i\)小的点，与后面\(n - i\)个点没有任何关联，所以后面\(n - i\)个点可以任意排布
这样我们就可以分治\(NTT\)在\(O(nlog^2n)\)的时间内预处理出\(f[i]\)
然后询问的时候根据乘法原理计算即可
时间复杂度\(O(nlog^2n + Tn)\)
如果您常数比较大，就需要优化一下，比如循环展开大法好
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define res register
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
const int G = 10,P = 786433;
int R[maxn],c[maxn],w[maxn];
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
void NTT(int* a,int n,int f){
for (res int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (res int i = 1; i < n; i <<= 1){
int gn = w[i];
for (res int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (res int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k],y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
}
}
}
if (f == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (res int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int fac[maxn],fv[maxn],N = 100000;
int f[maxn],A[maxn],B[maxn];
void solve(int l,int r){
if (l == r){f[l] = ((fac[l] - f[l]) % P + P) % P; return;}
int mid = l + r >> 1;
solve(l,mid);
int n,m,L = 0; n = mid - l;
for (res int i = 0; i <= n; i += 4){
A[i] = f[i + l]; A[i + 1] = f[i + l + 1];
A[i + 2] = f[i + l + 2]; A[i + 3] = f[i + l + 3];
}
n = r - l - 1;
for (res int i = 0; i <= n; i += 4){
B[i] = fac[i + 1]; B[i + 1] = fac[i + 2];
B[i + 2] = fac[i + 3]; B[i + 3] = fac[i + 4];
}
m = mid - (l << 1) + r - 1; n = 1;
while (n <= m) n <<= 1,L++;
for (res int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (res int i = mid - l + 1; i < n; i += 4){
A[i] = A[i + 1] = A[i + 2] = A[i + 3] = 0;
}
for (res int i = r - l + 1; i < n; i += 4){
B[i] = B[i + 1] = B[i + 2] = B[i + 3] = 0;
}
NTT(A,n,1); NTT(B,n,1);
for (res int i = 0; i < n; i += 4){
A[i] = 1ll * A[i] * B[i] % P;
A[i + 1] = 1ll * A[i + 1] * B[i + 1] % P;
A[i + 2] = 1ll * A[i + 2] * B[i + 2] % P;
A[i + 3] = 1ll * A[i + 3] * B[i + 3] % P;
}
NTT(A,n,-1);
for (res int i = mid - l; i <= r - l - 1; i++)
f[i + l + 1] = (f[i + l + 1] + A[i]) % P;
solve(mid + 1,r);
}
void init(){
fac[0] = 1;
for (res int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % P;
fv[N] = qpow(fac[N],P - 2); fv[0] = 1;
for (res int i = N - 1; i; i--)
fv[i] = 1ll * fv[i + 1] * (i + 1) % P;
for (res int i = 1; i < maxn; i <<= 1)
w[i] = qpow(G,(P - 1) / (i << 1));
f[0] = 1;
solve(1,N);
REP(i,N) f[i] = (f[i] + P) % P;
//REP(i,11) printf("%d ",f[i]); puts("");
}
int n,m,scc[maxn],vis[maxn];
int main(){
init();
int T = read();
while (T--){
n = read(); m = read(); int ans = 1,x;
REP(i,m) vis[i] = false;
REP(i,m){
x = read();
REP(j,x) scc[read()] = i;
ans = 1ll * ans * f[x] % P;
}
int flag = true;
for (res int i = 1; i <= n; i++){
if (i > 1 && scc[i] != scc[i - 1] && vis[scc[i]]){
flag = false; break;
}
vis[scc[i]] = true;
}
flag ? printf("%d\n",ans) : puts("0");
}
return 0;
}
