P4173

数学 #dfs

枚举一棵子树断开的边，标记这条边一个子树内的节点，在另一棵树上枚举断开的边，统计子树大小和与第一棵树匹配的点的个数

设第一棵树大小为 \(siz\)，第二棵树大小为 \(tot\) ，其中匹配的点数为 \(cnt\)

那么这一对边的 \(S(e_1,e_2)=max(cnt,tot-cnt,siz-cnt,n-tot-siz+cnt)\)

对于每一对边统计答案即可

时间复杂度 \(\mathcal{O}(n^2)\)

// Author: xiaruize
#ifndef ONLINE_JUDGE
bool start_of_memory_use;
#endif
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
	if (a > b)
		return a;
	return b;
}
int min(int a, int b)
{
	if (a < b)
		return a;
	return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 4e3 + 10;

int n;
vector<int> g[N], f[N];
int fag[N], faf[N];
bool vis[N];
int cntg = 0;
int cnt[N], siz[N];

void add(int x, int fa)
{
	// cerr << "add" << x << endl;
	cntg++;
	vis[x] = true;
	for (auto v : g[x])
		if (v != fa)
			add(v, x);
}

int res = 0;
void calc(int x, int fa)
{
	siz[x] = 1;
	cnt[x] = vis[x];
	for (auto v : f[x])
	{
		if (v == fa)
			continue;
		calc(v, x);
		cnt[x] += cnt[v];
		siz[x] += siz[v];
	}
	if (!x)
		return;
	int tmp = max({cnt[x], cntg - cnt[x], siz[x] - cnt[x], n + 1 - cntg - siz[x] + cnt[x]});
	// cerr << x << ' ' << cnt[x] << ' ' << siz[x] << ' ' << cntg << ' ' << tmp << endl;
	res += tmp * tmp;
}

void dfs(int x, int fa)
{
	// cerr << x << endl;
	for (auto v : g[x])
	{
		if (v != fa)
		{
			dfs(v, x);
		}
	}
	if (!x)
		return;
	// cerr << x << ' ' << fa << endl;
	cntg = 0;
	mms(vis, 0);
	add(x, fa);
	// cerr << cntg << endl;
	mms(cnt, 0);
	mms(siz, 0);
	calc(0, 0);
	// cerr << x << ' ' << res << endl;
}

void solve()
{
	cin >> n;
	rep(i, 0, n - 1)
	{
		int x;
		cin >> x;
		g[i].push_back(x);
		g[x].push_back(i);
	}
	cin >> n;
	rep(i, 0, n - 1)
	{
		int x;
		cin >> x;
		f[i].push_back(x);
		f[x].push_back(i);
	}
	// cerr << "flag" << endl;
	dfs(0, 0);
	cout << res << endl;
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}