[CodeForces 156D] Clues（Prufer 序列 + 多项式定理） | 错题本

题目

[CodeForces 156D] Clues

分析

记联通块个数为 $k$，枚举每个联通块的度数 $d_i$（ $\sum_{i = 1}^{k} d_i = 2(k - 1)$，每条边的贡献为 $2$，因此度数和为 $2(k - 1)$），用重排列计算方案数，并且对于一个联通块，取里面一个点的方案数为联通块的大小 $s_i$，可得答案为 $\begin{aligned} & \sum_{\sum_{i = 1}^{k} d_i = 2(k - 1),\ d_i \geq 1} \left( \frac{\left( \sum_{i = 1}^{k}(d_i - 1) \right) !}{\prod_{i = 1}^{k} ((d_i - 1)!)} \prod_{i = 1}^{k} {s_i}^{d_i} \right) \\ =& \sum_{\sum_{i = 1}^{k} d_i = 2(k - 1),\ d_i \geq 1} \left( \frac{(k - 2)!}{\prod_{i = 1}^{k} ((d_i - 1)!)} \prod_{i = 1}^{k} {s_i}^{d_i} \right) \end{aligned}$ 令 $p_i = d_i - 1$，则原式变为 $\sum_{\sum_{i = 1}^{k} p_i = k - 2,\ p_i \geq 0} \left( \frac{(k - 2)!}{\prod_{i = 1}^{k} (p_i!)} \prod_{i = 1}^{k} {s_i}^{p_i + 1} \right)$ 又由多项式定理 $\left( \sum_{i = 1}^{k} s_i \right)^{k - 2} = \sum_{\sum_{i = 1}^{k} p_i = k - 2, p_i \geq 0} \left( \frac{(k - 2)!}{\prod_{i = 1}^{k} (p_i!)} \prod_{i = 1}^{k} {s_i}^{p_i} \right)$ 因此，原式即为 $\left( \sum_{i = 1}^{k} s_i \right)^{k - 2} \prod_{i = 1}^{k} s_i = n^{k - 2} \prod_{i = 1}^{k} s_i$ 注意特判 $k = 1$ 的情况，并且要输出 1 % MOD。

错因

• 并查集都能写错；
• 没有特判 $k = 1$ 的情况；
• 加上特判的时候只 puts("1")，而不是 printf("%d", 1 % MOD)；
• 并查集的 size 可以直接合并。

代码

#include <bits/stdc++.h>

const int MAXN = 100000;

int N, MOD, M, K;

int Fa[MAXN + 5], Size[MAXN + 5];

int Find(int x) {
    return (Fa[x] == x) ? x : (Fa[x] = Find(Fa[x]));
}

int Pow(int x, int y) {
    int ret = 1;
    while (y) {
        if (y & 1)
            ret = (long long)ret * x % MOD;
        x = (long long)x * x % MOD;
        y >>= 1;
    }
    return ret;
}

int main() {
    scanf("%d%d%d", &N, &M, &MOD);
    for (int i = 1; i <= N; i++)
        Fa[i] = i, Size[i] = 1;
    for (int i = 1; i <= M; i++) {
        int u, v; scanf("%d%d", &u, &v);
        int x = Find(u), y = Find(v);
        if (x != y)
            Fa[x] = y, Size[y] += Size[x];
    }
    int Ans = 1;
    for (int i = 1; i <= N; i++) {
        Find(Fa[i]);
        if (Fa[i] == i)
            K++, Ans = (long long)Ans * Size[i] % MOD;
    }
    if (K == 1)
        return printf("%d", 1 % MOD), 0;
    Ans = (long long)Ans * Pow(N, K - 2) % MOD;
    printf("%d", Ans);
    return 0;
}