hdu5279 YJC plays Minecraft 【分治NTT】
题目链接
题解
给出若干个完全图，然后完全图之间首尾相连并成环，要求删边使得两点之间路径数不超过\(1\)，求方案数
容易想到各个完全图是独立的，每个完全图要删成一个森林，其实就是询问\(n\)个点有标号森林的个数
设\(f[i]\)表示\(i\)个点有标号森林的个数
枚举第一个点所在树大小，我们只需求出\(n\)个点有多少种树，由\(purfer\)序容易知道是\(n^{n - 2}\)
那么有
\[f[n] = \sum\limits_{i = 1}^{n} {n - 1 \choose i - 1}i^{i - 2}f[n - i]
\]
化简一下：
\[f[n] = (n - 1)!\sum\limits_{i = 1}^{n}\frac{i^{i - 2}}{(i - 1)!} \times \frac{f[n - i]}{(n - i)!}
\]
分治\(NTT\)即可
每个完全图的方案是\(f[a[i]]\)，中间相连的\(n\)条边有\(2^n\)种方案，由乘法原理乘起来即可
但是这样求出来的不是答案，会多算一类情况：
每个完全图的\(1\)和\(a_i\)相通且所有中介边存在
所以我们还需要计算\(g[i]\)表示\(i\)个点的森林，\(1\)和\(i\)点在同一棵树内的方案数
显然
\[g[n] = \sum\limits_{i = 2}^{n} {n - 2 \choose i - 2}i^{i - 2}f[n - i]
\]
化简得
\[g[n] = (n - 2)!\sum\limits_{i = 2}^{n} \frac{i^{i - 2}}{(i - 2)!} \times \frac{f[n - i]}{(n - i)!}
\]
\(NTT\)即可
最后答案减去\(g[a[i]]\)的乘积即可
复杂度\(O(nlog^2n)\)
#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 (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
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 = 3,P = 998244353;
int R[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 (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (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 (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int f[maxn],g[maxn],fac[maxn],fv[maxn],p[maxn],N = 100005;
int A[maxn],B[maxn];
void solve(int l,int r){
if (l == r){
if (l > 0) f[l] = 1ll * f[l] * fac[l - 1] % P;
return;
}
int mid = l + r >> 1;
solve(l,mid);
int n,m,L;
m = mid - l + 1;
for (int i = 0; i < m; i++) A[i] = 1ll * f[l + i] * fv[l + i] % P;
m = r - l;
for (int i = 0; i < m; i++) B[i] = 1ll * p[i + 1] * fv[i] % P;
n = 1; L = 0; m = mid + r - (l << 1) - 1;
while (n <= m) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (int i = mid - l + 1; i < n; i++) A[i] = 0;
for (int i = r - l; i < n; i++) B[i] = 0;
NTT(A,n,1); NTT(B,n,1);
for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;
NTT(A,n,-1);
for (int i = mid - l,j = mid + 1; j <= r; i++,j++){
f[j] = (f[j] + A[i]) % P;
}
solve(mid + 1,r);
}
int b[maxn];
inline int C(int n,int m){
if (m > n) return 0;
return 1ll * fac[n] * fv[m] % P * fv[n - m] % P;
}
void work(){
fac[0] = p[0] = p[1] = 1;
for (int i = 1; i <= N + 2; i++)
fac[i] = 1ll * fac[i - 1] * i % P;
for (int i = 2; i <= N + 2; i++)
p[i] = qpow(i,i - 2);
fv[N + 2] = qpow(fac[N + 2],P - 2); fv[0] = 1;
for (int i = N + 1; i; i--)
fv[i] = 1ll * fv[i + 1] * (i + 1) % P;
f[0] = 1;
solve(0,N);
A[0] = A[1] = 0;
for (int i = 2; i <= N; i++) A[i] = 1ll * p[i] * fv[i - 2] % P;
for (int i = 0; i <= N; i++) B[i] = 1ll * f[i] * fv[i] % P;
int n = 1,L = 0;
while (n <= (N << 1)) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (int i = N + 1; i < n; i++) A[i] = 0;
for (int i = N + 1; i < n; i++) B[i] = 0;
NTT(A,n,1); NTT(B,n,1);
for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;
NTT(A,n,-1);
for (int i = 2; i <= N; i++) g[i] = 1ll * A[i] * fac[i - 2] % P;
g[1] = 1;
}
int n,a[maxn],ans,ans2;
int main(){
work();
//REP(i,100) printf("%d ",f[i]); puts("");
//REP(i,100) printf("%d ",g[i]); puts("");
int T = read();
while (T--){
n = read();
REP(i,n) a[i] = read();
ans = qpow(2,n);
REP(i,n) ans = 1ll * ans * f[a[i]] % P;
ans2 = 1;
REP(i,n) ans2 = 1ll * ans2 * g[a[i]] % P;
ans = ((ans - ans2) % P + P) % P;
printf("%d\n",ans);
}
return 0;
}
