BZOJ 3456: 城市規劃 與 多項式求逆算法介紹(多項式求逆, dp)

題面

求有 \(n\) 個點的無向有標號連通圖個數 . \((1 \le n \le 1.3 * 10^5)\)

題解

首先考慮 dp ... 直接算可行的方案數 , 容易算重複 .

咱們用總方案數減去不可行的方案數就好了 (容斥)

令 \(f_i\) 爲有 \(i\) 個點的無向有標號連通圖個數 .

考慮 \(1\) 號點的聯通塊大小 , 聯通塊外的點之間邊任意 但 不能與 \(1\) 有間接聯繫 .

那麼就有

\[\displaystyle f_i = 2^{\binom i 2} - \sum_{j=1}^{i-1} f_j \times \binom {i-1}{j-1} \times 2^{\binom{i-j}{2}}\]

這個能夠直接用 CDQ分治FFT 求 , 但我不會 ...

那認真考慮一下這個式子 qwq

先展開組合數

\[\displaystyle f_i=2^{\binom{i}{2}} - \sum_{j=1}^{i-1} f_j \times \frac{(i-1)!}{(j-1)!(i-j)!} \times 2^{\binom {i-j}{2}}\]

除去 \((i-1)!\)

\[\displaystyle \frac{f_i}{(i-1)!} = \frac{2^{\binom i 2}}{(i-1)!} - \sum_{j=1}^{i-1} \frac{f_j\times2^{\binom{i-j}{2}}}{(j-1)!(i-j)!}\]

移項合併

\[\displaystyle \sum_{j=1}^{i} \frac{f_j \times 2^{\binom {i-j}{2}}}{(j-1)!\times(i-j)!} = \frac{2^{\binom i 2}}{(i-1)!}\]

左邊咱們觀察一下不難發現是一個卷積形式 .

令 \(\displaystyle A=\sum_{i=1}^{n} \frac{f_i}{(i-1)!} x^i\)

\(\displaystyle B=\sum_{i=0}^{n} \frac{2^{\binom {i}{2}}}{i!} x^i\)

\(\displaystyle C = \sum_{i=1}^{n} \frac{2^{\binom{i}{2}}}{(i-1)!} x^i\)

就有 \(A * B = C \Rightarrow A=C * B^{-1}\)

咱們只須要求出 \(B\) 在 \(x^n\) 下的逆元就好了 qwq

怎麼求呢 :

多項式求逆 :

若是 \(B\) 爲 \(A\) 在 \(x^n\) 意義下的逆元 , 那麼數學表達就是 :

\[A * B \equiv 1 \pmod {x^n}\]

假設咱們已經求出了\(B\) 在 \(\displaystyle x^{\frac{n}{2}}\) 的逆元 \(B'\) . (咱們經常令 \(n\) 爲 \(2^k\) )

\(\displaystyle A * B' \equiv 1 \pmod {x^\frac{n}{2}} \tag{1}\)

咱們以前那個 \(x^{\frac{n}{2}}\) 意義下也成立 ... 就有

\[\displaystyle A * B \equiv 1 \pmod {x^{\frac{n}{2}}} \tag{2}\]

讓 \((2)-(1)\) 就有

\[B - B'\equiv 0 \pmod{x^{\frac n 2}}\]

而後把它左右平方一下 (此處 \(x^\frac{n}{2}\) 也可平方成 \(x^n\) )

\[B^2 -2BB' +B'^2 \equiv 0 \pmod {x^{n}}\]

乘上一個 \(A\) 就獲得

\[B \equiv 2B' - A * B'^2 \pmod {x^n}\]

而後遞歸求解就好了 .

那麼複雜度就是 \(T(n) = T(\frac{n}{2}) + O(n \log n) =O(n \log n)\) ...

代碼在這道題程序中有 ...

代碼

/**************************************************************
    Problem: 3456
    User: DOFY
    Language: C++
    Result: Accepted
    Time:11956 ms
    Memory:107796 kb
****************************************************************/
 
#include <bits/stdc++.h>
#define For(i, l, r) for(register int i = (l), i##end = (int)(r); i <= i##end; ++i)
#define Fordown(i, r, l) for(register int i = (r), i##end = (int)(l); i >= i##end; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;
 
typedef long long ll;
inline bool chkmin(int &a, int b) {return b < a ? a = b, 1 : 0;}
inline bool chkmax(int &a, int b) {return b > a ? a = b, 1 : 0;}
 
inline int read() {
    int x = 0, fh = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar()) if (ch == '-') fh = -1;
    for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ 48);
    return x * fh;
}
 
void File() {
#ifdef zjp_shadow
    freopen ("3456.in", "r", stdin);
    freopen ("3456.out", "w", stdout);
#endif
}
 
const int N = (1 << 21) + 5, Mod = 1004535809;
 
ll fpm(ll x, int power) {
    ll res = 1;
    for (; power; power >>= 1, (x *= x) %= Mod)
        if (power & 1) (res *= x) %= Mod;
    return res;
}
 
ll fac[N], ifac[N];
 
inline ll C(int n, int m) {
    if (n < 0 || m < 0 || n < m) return 0;
    return fac[n] * ifac[m] % Mod * ifac[n - m] % Mod;
}
 
void Init(int maxn) {
    fac[0] = ifac[0] = 1;
    For (i, 1, maxn) fac[i] = fac[i - 1] * i % Mod;
    ifac[maxn] = fpm(fac[maxn], Mod - 2);
    Fordown (i, maxn - 1, 1) ifac[i] = ifac[i + 1] * (i + 1) % Mod;
}
 
inline int Add(int a, int b) { return ((a += b) >= Mod) ? a - Mod : a; }
struct Number_Theory_Transform {
    int pow3[N], invpow3[N];
    inline void Init(int maxn) {
        for (int i = 2; i <= maxn; i <<= 1)
            pow3[i] = fpm(3, (Mod - 1) / i), invpow3[i] = fpm(pow3[i], Mod - 2);
    }
 
    int n, rev[N];
    inline void Get_Rev() {
        int cnt = 0; for (int i = 1; i < n; i <<= 1) ++ cnt;
        For (i, 0, n - 1) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));
    }
 
    void NTT(int P[], int opt) {
        For (i, 0, n - 1) if (i < rev[i]) swap(P[i], P[rev[i]]);
        for (int i = 2; i <= n; i <<= 1) {
            int p = i >> 1, Wi = opt == 1 ? pow3[i] : invpow3[i];
            for (int j = 0; j < n; j += i)
                for (int k = 0, x = 1; k < p; ++ k, x = 1ll * x * Wi % Mod) {
                    int u = P[j + k], v = 1ll * x * P[j + k + p] % Mod;
                    P[j + k] = Add(u, v);
                    P[j + k + p] = Add(u, Mod - v);
                }
        }
        if (opt == -1) {
            int invn = fpm(n, Mod - 2);
            For (i, 0, n - 1) P[i] = 1ll * P[i] * invn % Mod;
        }
    }
 
    int A[N], B[N];
    void Mult(int a[], int b[], int c[], int len) {
        for (n = 1; n <= len; n <<= 1); Get_Rev();
        For (i, 0, n - 1) A[i] = a[i], B[i] = b[i];
 
        NTT(A, 1); NTT(B, 1);
        For (i, 0, n - 1) A[i] = 1ll * A[i] * B[i] % Mod;
        NTT(A, -1);
 
        For (i, 0, n - 1) c[i] = A[i];
    }
 
    void Get_Inv(int a[], int b[], int len) {
        if (len == 1) { b[0] = fpm(a[0], Mod - 2); return ; }
        Get_Inv(a, b, len >> 1);
 
        n = len << 1; Get_Rev();
        For (i, 0, len - 1) A[i] = a[i], B[i] = b[i];
 
        NTT(A, 1); NTT(B, 1);
        For (i, 0, n - 1) A[i] = 1ll * A[i] * B[i] % Mod * B[i] % Mod;
        NTT(A, - 1);
 
        For (i, 0, len - 1)
            b[i] = Add(Add(b[i], b[i]), Mod - A[i]);
    }
} T;
 
int a[N], b[N], c[N], tmp[N], n;
 
int Edge(int x, int nowmod) { return 1ll * x * (x - 1) / 2 % nowmod; }
 
int main () {
    File();
    n = read();
    T.Init(1 << 20); Init(n);
 
    For (i, 0, n)
        b[i] = fpm(2, Edge(i, Mod - 1)) * ifac[i] % Mod;
 
    For (i, 0, n)
        c[i] = fpm(2, Edge(i, Mod - 1)) * ifac[i - 1] % Mod;
 
    int len = 1; for (; len <= n; len <<= 1);
    T.Get_Inv(b, tmp, len); T.Mult(tmp, c, a, len);
 
    printf ("%lld\n", 1ll * a[n] * fac[n - 1] % Mod);
 
    return 0;
}