给定一棵nn个节点的树。

对于每个点单独考虑，删掉它，你可以把一个节点的父亲改成另一个节点（只能做一次），使森林中最大的树sizesize最小。对每个点求出这个答案

n≤105n\le 10^5

CF768G

Solution

不难发现，在点 xx 处计算答案时，会从最大的联通块中选择一棵子树接到最小的联通块上

二分答案, 转化成区间存在性问题, 直接用multiset在每个点上维护子树内所有点的 sizesize 信息

此外还需要单独考虑一下子树外联通块的情况，并且这个点到根的这条链上的信息也要单独维护，因为它们的sizesize会减掉当前点的sizesize

树上启发式合并维护即可

具体细节见代码

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }

inline void proc_status()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) <<endl;
}

template <typename T> T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

const int Maxn = 1e5 + 100;

int N;
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1];
int size[Maxn], son[Maxn], snd[Maxn], min_size[Maxn];
int Ans[Maxn];

multiset <int> S[4];
//S[0]: anc, S[1]: heavy, S[2]: light, S[3]: other

inline void add_edge (int x, int y) { To[++e] = y; Next[e] = Begin[x]; Begin[x] = e; }

inline void dfs_pre (int x)
{
	size[x] = 1, min_size[x] = 0x3f3f3f3f;

	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		dfs_pre(y);
		size[x] += size[y];
		if (size[y] > size[son[x]]) snd[x] = size[son[x]], son[x] = y;
		else if (size[y] > snd[x]) snd[x] = size[y];
		Chkmin(min_size[x], size[y]);
	}

	if (size[x] != N) Chkmin(min_size[x], N - size[x]);
}

inline void Modify (int x, int val, int k)
{
	if (val == 1) S[k].insert(x), S[3].erase(S[3].find(x));
	else S[k].erase(S[k].find(x)), S[3].insert(x);
}

inline void Add (int x, int k)
{
	Modify (size[x], 1, k);
	for (int i = Begin[x]; i; i = Next[i]) Add (To[i], k);
}

inline void Erase (int x, int k)
{
	Modify (size[x], -1, k);
	for (int i = Begin[x]; i; i = Next[i]) Erase (To[i], k);
}

inline int Find (int Min, int Max, int l, int r, int k, int delta)
{
	if (Min == Max && l <= Min && Min <= r) return Min;

	int ans = 0x3f3f3f3f;
	while (l <= r)
	{
		int mid = l + r >> 1;
		multiset <int> :: iterator x = S[k].lower_bound(Max - mid + delta);
		if (x != S[k].end() && (*x) <= mid - Min + delta) r = mid - 1, ans = mid;
		else l = mid + 1;
	}
	return ans;
}

inline void Calc (int x)
{
	if (!son[x]) { Ans[x] = N - size[x]; return; }
	if (size[x] == N || size[son[x]] > N - size[x]) // max subtree is heavy
		Ans[x] = Find (min_size[x], size[son[x]], max(N - size[x], snd[x]), size[son[x]], 1, 0);
	else // root is in the max subtree
		Ans[x] = min(Find (min_size[x], N - size[x], size[son[x]], N - size[x], 0, size[x]), 
					 Find (min_size[x], N - size[x], size[son[x]], N - size[x], 3, 0));
}

multiset <int> :: iterator it, Save[Maxn];

inline void dfs (int x, int flag)
{
	Modify (size[x], 1, 0);

	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (y != son[x]) dfs(y, 0);
	}

	if (son[x]) dfs(son[x], 1);

	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (y != son[x]) Add (y, 2);
	}

	S[0].erase(S[0].find(size[x]));
	Calc(x); 
	S[3].insert(size[x]);

	if (flag)
	{
		Modify(size[x], 1, 1);
		int top = 0;
		for (it = S[2].begin(); it != S[2].end(); ++it) S[1].insert(*it), Save[++top] = it;
		for (int i = 1; i <= top; ++i) S[2].erase(Save[i]);
	}
	else
	{
		for (int i = Begin[x]; i; i = Next[i])
		{
			int y = To[i];
			if (y != son[x]) Erase (y, 2);
			else Erase (y, 1);
		}
	}
}

int root;

inline void Solve ()
{
	dfs_pre (root);
	for (int i = 1; i <= N; ++i) S[3].insert(size[i]);

	dfs(root, 0);

	for (int i = 1; i <= N; ++i) printf("%d\n", Ans[i]);
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i)
	{
		int x = read<int>(), y = read<int>();
		if (!x) { root = y; continue; }
		add_edge (x, y);
	}
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("G.in", "r", stdin);
	freopen("G.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}