BZOJ3625: 小朋友和二叉树

二叉树计数与生成函数

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 60 阅读

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原文链接：http://www.cnblogs.com/cjoieryl/p/10039732.html

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#数据结构与算法

本文介绍了一种使用生成函数求解特定条件下二叉树计数问题的方法，通过构造复杂的数学公式和算法实现，重点在于如何利用生成函数和快速傅立叶变换(FFT)的逆变换来高效计算二叉树的数量。

传送门

Sol

设 \(f_x\) 表示权值为 \(x\) 的二叉树的个数
设 \(s_x\) 表示是否有 \(x\) 这种权值可以选择
那么
\[f_n=\sum_{i=0}^{n}\sum_{j=0}^{n-i}f_jf_{n-i-j}s_i\]
构造
\[F(x)=\sum_{i=0}f_ix^i\]
\[S(x)=\sum_{i=0}s_ix^i\]
由于 \(s_0=0,f_0=1\)
那么
\(F^2(x)S(x)=F(x)-1\)
所以可以求得
\[F(x)=\frac{1\pm \sqrt{1-4S(x)}}{2S(x)}\]
由于 \(F(0)=1,S(0)=0\) 所以
\[F(x)=\frac{1- \sqrt{1-4S(x)}}{2S(x)}\]
正是因为 \(S(0)=0\) 没有办法求逆
所以化简得到
\[F(x)=\frac{2}{1+ \sqrt{1-4S(x)}}\]
开根(常数项是 \(1\) 所以不用二次剩余)+求逆即可

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(4e5 + 5);
const int mod(998244353);
const int inv2(499122177);

inline int Pow(ll x, int y) {
    register ll ret = 1;
    for (; y; y >>= 1, x = x * x % mod)
        if (y & 1) ret = ret * x % mod;
    return ret;
}

inline void Inc(int &x, int y) {
    if ((x += y) >= mod) x -= mod;
}

int a[maxn], b[maxn], c[maxn], w[2][maxn], deg, r[maxn], l;

inline void Init(int n) {
    register int i, k, wn, iwn;
    for (deg = 1, l = 0; deg < n; deg <<= 1) ++l;
    for (i = 0; i < deg; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
    for (i = 1; i < deg; i <<= 1) {
        w[0][0] = w[1][0] = 1;
        wn = Pow(3, (mod - 1) / (i << 1)), iwn = Pow(wn, mod - 2);
        for (k = 1; k < i; ++k) {
            w[0][deg / i * k] = 1LL * w[0][deg / i * (k - 1)] * wn % mod;
            w[1][deg / i * k] = 1LL * w[1][deg / i * (k - 1)] * iwn % mod;
        }
    }
}

inline void NTT(int *p, int opt) {
    register int i, j, k, t, wn, x, y;
    for (i = 0; i < deg; ++i) if (r[i] < i) swap(p[r[i]], p[i]);
    for (i = 1; i < deg; i <<= 1)
        for(t = i << 1, j = 0; j < deg; j += t)
            for (k = 0; k < i; ++k) {
                wn = w[opt == -1][deg / i * k];
                x = p[j + k], y = 1LL * wn * p[i + j + k] % mod;
                p[j + k] = x + y, p[i + j + k] = x - y;
                if (p[j + k] >= mod) p[j + k] -= mod;
                if (p[i + j + k] < 0) p[i + j + k] += mod;
            }
    if (opt == -1) {
        wn = Pow(deg, mod - 2);
        for (i = 0; i < deg; ++i) p[i] = 1LL * p[i] * wn % mod;
    }
}

int n, m, f[maxn], g[maxn], s[maxn];

void Inv(int *p, int *q, int len) {
    if (len == 1) {
        q[0] = Pow(p[0], mod - 2);
        return;
    }
    Inv(p, q, len >> 1);
    register int i, tmp = len << 1;
    for (i = 0; i < len; ++i) a[i] = p[i], b[i] = q[i];
    Init(tmp), NTT(a, 1), NTT(b, 1);
    for (i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * b[i] % mod * b[i] % mod;
    NTT(a, -1);
    for (i = 0; i < len; ++i) q[i] = (2LL * q[i] + mod - a[i]) % mod;
    for (i = 0; i < tmp; ++i) a[i] = b[i] = 0;
}

void Sqrt(int *p, int *q, int len) {
    if (len == 1) {
        q[0] = sqrt(p[0]);
        return;
    }
    Sqrt(p, q, len >> 1), Inv(q, c, len);   
    register int i, tmp = len << 1;
    for (i = 0; i < len; ++i) a[i] = p[i];
    Init(tmp), NTT(a, 1), NTT(c, 1);
    for (i = 0; i < tmp; ++i) a[i] = 1LL * a[i] * c[i] % mod;
    NTT(a, -1);
    for (i = 0; i < len; ++i) q[i] = 1LL * (q[i] + a[i]) % mod * inv2 % mod;
    for (i = 0; i < tmp; ++i) a[i] = c[i] = 0;
}

int main() {
    register int i, len, v;
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; ++i) scanf("%d", &v), ++s[v];
    for (len = 1; len <= m; len <<= 1);
    for (i = 1; i <= m; ++i) s[i] = 1LL * s[i] * 4 % mod, s[i] = mod - s[i];
    s[0] = 1, Sqrt(s, f, len), Inc(f[0], 1), Inv(f, g, len);
    for (i = 0; i < len; ++i) g[i] = 2LL * g[i] % mod;
    for (i = 1; i <= m; ++i) printf("%d\n", g[i]);
    return 0;
}

转载于:https://www.cnblogs.com/cjoieryl/p/10039732.html