[ARC150D] Removing Gacha 题解

合集 - 题解(65)

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49.[ARC150D] Removing Gacha 题解2023-10-03

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收起

题意

给定一棵由 $N$ 个节点组成的树，每个节点有黑白两种颜色。定义一个节点 $u$ 为好的当且仅当路径 $1\leftrightarrow u$ 上的节点均为黑色的，反之为坏的。初始情况下所有点均为白色。

定义一次操作为选取一个坏的节点并将其染黑，求将全部节点均染为黑色的期望操作次数，对 $998244353$ 取模。

（$2\le N\le 2\times {10}^5$）。

题解

发现所有节点均为黑色是所有节点均是好的的充要条件。所以我们可以将问题转化为求将所有节点变为好的的期望次数。

设 $E(X_i)$ 代表直至第 $i$ 期望变为好的，第 $i$ 个点期望被选中的次数，那么我们要求的所有节点的该值的和。

下面考虑如何计算该值，我们可以考虑先计算将第 $i$ 个点变为好的的期望选择路径 $1\leftrightarrow u$ 上的节点的次数。可以发现该问题与如下问题等价：

有 $n$ 种无限个的物品，每次随机等概率选择一种物品，求使得所有物品均被选择的期望次数。

可以考虑设当前选择了 $k$ 种物品，那么下一次选择到新物品的概率为 ${\displaystyle \frac{n-k}{n}}$，那么选择到新物品的期望次数即为 ${\displaystyle \frac{n}{n-k}}$，那么总的期望次数即为

$$\sum _{k=0}^{n-1}{\displaystyle \frac{n}{n-k}}=n\sum _{i=1}^n{\displaystyle \frac{1}{i}}$$

发现在这个问题中不同种类的物品等价，那么我们可以得出

$$E(X_i)={\displaystyle \frac{{\mathrm{depth}}_u\sum _{i=1}^{{\mathrm{depth}}_u}{\displaystyle \frac{1}{i}}}{{\mathrm{depth}}_u}}=\sum _{i=1}^{{\mathrm{depth}}_u}{\displaystyle \frac{1}{i}}$$

$\mathcal{O}(N)$ 求出即可。值得一提，由于本题规定每个节点的父亲节点编号小于该节点编号，所以无需显式建树。

Code

#include <bits/stdc++.h>

typedef long long valueType;

typedef std::vector<valueType> ValueVector;

constexpr valueType MOD = 998244353;

template<typename T1, typename T2, typename T3 = valueType>
void Inc(T1 &a, T2 b, const T3 &mod = MOD) {
    a = a + b;

    if (a >= mod)
        a -= mod;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 sum(T1 a, T2 b, const T3 &mod = MOD) {
    return a + b >= mod ? a + b - mod : a + b;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 mul(T1 a, T2 b, const T3 &mod = MOD) {
    return (long long) a * b % mod;
}

template<typename T1, typename T2, typename T3 = valueType>
void Mul(T1 &a, T2 b, const T3 &mod = MOD) {
    a = (long long) a * b % mod;
}

template<typename T1, typename T2, typename T3 = valueType>
T1 pow(T1 a, T2 b, const T3 &mod = MOD) {
    T1 result = 1;

    while (b > 0) {
        if (b & 1)
            Mul(result, a, mod);

        Mul(a, a, mod);
        b = b >> 1;
    }

    return result;
}

class Inverse {
private:
    valueType N;
    ValueVector data;

public:
    explicit Inverse(valueType n = 0) : N(n), data(n + 1) {
        data[1] = 1;

        for (valueType i = 2; i <= N; ++i)
            data[i] = mul(MOD - MOD / i, data[MOD % i]);
    }

    valueType operator()(valueType n) const {
        return data[n];
    }
};

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);

    valueType N;

    std::cin >> N;

    Inverse const Inv(N);

    ValueVector depth(N + 1);

    depth[1] = 1;

    ValueVector H(N + 1, 0);

    for (valueType i = 1; i <= N; ++i)
        H[i] = sum(H[i - 1], Inv(i));

    valueType ans = H[1];

    for (valueType i = 2; i <= N; ++i) {
        valueType father;

        std::cin >> father;

        depth[i] = depth[father] + 1;

        Inc(ans, H[depth[i]]);
    }

    std::cout << ans << std::endl;

    return 0;
}