rt，本题按照如下方法哈希可过：

选取一棵树的每个点为根，搜出所有点的深度和子树大小，然后暴力哈希。

具体见下面一个 AC 的暴力代码：

#include<bits/stdc++.h>
using namespace std;
const int mod1 = 998244353, mod2 = 998244352;
int t, n, rt, fst, f[55], siz[55];
vector<int> G[55];
struct node{
	long long h1, h2;
	bool operator < (node b)const{
		return this->h1 < b.h1;
	}
};
long long h1[55], h2[55];
int pr[105], cnt;
bool vis[105];
void sieve(int lim){
	for (int i = 2; i <= lim; i++){
		if (!vis[i]){
			pr[++cnt] = i;
			vis[i] = 1;
		}
		for (int j = 1; j <= cnt; j++){
			if (i * pr[j] > lim)
				break;
			vis[i * pr[j]] = 1;
			if (i % pr[j] == 0)
				break;
		}
	}
}
void dfs(int u, int fa, int dep){
	f[u] = dep;
	if (dep > f[fst])
		fst = u;
	for (auto v : G[u]){
		if (v == fa)
			continue;
		dfs(v, u, dep + 1);
	}
}
void dfs2(int u, int fa){
	for (auto v : G[u])
		if (v != fa){
			dfs2(v, u);
			siz[u] += siz[v];
		}
	siz[u]++;
}
long long dfs3(int u, int fa, const int mod){
	long long res = f[u];
	for (auto v : G[u])
		if (v != fa)
			res = (res + dfs3(v, u, mod) * pr[siz[v]] % mod * pr[f[v]] % mod) % mod;
	return res;
}
void hsh(long long &h1, long long &h2){
    for (int i = 1; i <= n; i++){
        memset(siz, 0, sizeof siz);
        memset(f, 0, sizeof siz);
        dfs(i, 0, 0);
	    dfs2(i, 0);
	    h1 = (h1 + dfs3(i, 0, mod1)) % mod1;
	    h2 = (h2 + dfs3(i, 0, mod2)) % mod2;
    }
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	sieve(100);
	cin >> t;
	for (int iii = 1; iii <= t; iii++){
		cin >> n;
		for (int i = 1; i <= n; i++)
			G[i].clear();
		long long hsh1 = n, hsh2 = n;
		for (int i = 1; i <= n; i++){
			int fa;
			cin >> fa;
			if (!fa){
				rt = i;
				continue;
			}
			G[fa].push_back(i);
			G[i].push_back(fa);
		}
		hsh(hsh1, hsh2);
		if (hsh1 < hsh2)
			swap(hsh1, hsh2);
		h1[iii] = hsh1;
		h2[iii] = hsh2;
	//	cout << d1 << " " << d2 << " " << hsh1 << " " << hsh2 << "\n"; 
		int fd = iii;
		for (int j = iii - 1; j; j--)
			if (h1[j] == h1[iii] && h2[j] == h2[iii])
				fd = j;
		cout << fd << "\n";
	}
}