AtCoder Regular Contest 162 D Smallest Vertices

洛谷传送门

AtCoder 传送门

注意到，如果给定每个点的度数 \(d_i\)，那么根据 Prufer 序可得，合法本质不同无根树数量为 \(\frac{(n - 2)!}{\prod\limits_{i = 1}^n (d_i - 1)!}\)。

考虑拆贡献，钦定一个点是好点，在这样的基础上求本质不同无根树数量。

设钦定 \(u\) 为好点。那么 \(u\) 子树内所有点都在 \([u, n]\) 内。注意到求无根树的式子只关心树的大小和 \((d_i - 1)!\) 的乘积，并且合法当且仅当 \(\sum (d_i - 1) = n - 2\)，于是考虑背包，设 \(f_{i, j, k}\) 为 \([i, n]\) 中，被选的点度数和为 \(j\)，数量为 \(k\) 的方案数。

考虑完 \(u\) 子树后，把这些点全部删掉，\(u\) 作为叶子再参与接下来整棵树方案数的计算即可。

时间复杂度 \(O(n^3)\)。

code

// Problem: C - Mex Game on Tree
// Contest: AtCoder - AtCoder Regular Contest 162
// URL: https://atcoder.jp/contests/arc162/tasks/arc162_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1010;

int n, m, a[maxn], rnk[maxn], st[maxn], ed[maxn], times, sz[maxn];
bool flag, ff;
vector<int> G[maxn];
set<int> S[maxn];

void dfs(int u) {
	st[u] = ++times;
	rnk[times] = u;
	sz[u] = (a[u] == -1);
	for (int v : G[u]) {
		dfs(v);
		sz[u] += sz[v];
		for (int x : S[v]) {
			S[u].insert(x);
		}
	}
	ed[u] = times;
	if (a[u] != -1) {
		S[u].insert(a[u]);
	}
	if (sz[u] == 1) {
		set<int> T;
		for (int i = st[u]; i <= ed[u]; ++i) {
			int v = rnk[i];
			if (a[v] != -1) {
				T.insert(a[v]);
			}
		}
		int mex = 0;
		while (T.find(mex) != T.end()) {
			++mex;
		}
		if (mex == m) {
			ff = 1;
		} else if (mex > m) {
			return;
		}
		for (int i = st[u]; i <= ed[u]; ++i) {
			int v = rnk[i];
			if (a[v] == -1) {
				T.insert(mex);
				while (T.find(mex) != T.end()) {
					++mex;
				}
				if (mex == m) {
					ff = 1;
				}
				break;
			}
		}
	} else if (sz[u] == 0) {
		int mex = 0;
		while (S[u].find(mex) != S[u].end()) {
			++mex;
		}
		if (mex == m) {
			ff = 1;
		}
	}
}

void solve() {
	times = 0;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; ++i) {
		vector<int>().swap(G[i]);
		S[i].clear();
		rnk[i] = st[i] = ed[i] = sz[i] = 0;
	}
	for (int i = 2, p; i <= n; ++i) {
		scanf("%d", &p);
		G[p].pb(i);
	}
	for (int i = 1; i <= n; ++i) {
		scanf("%d", &a[i]);
	}
	ff = 0;
	dfs(1);
	if (ff) {
		puts("Alice");
		return;
	}
	puts("Bob");
}

int main() {
	int T = 1;
	scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}