给你一棵nn个节点的树

定义 f(x,y)=∑p∈subtree(y),p!=y[dep(p)≤dep(x)]f (x, y) = \sum_{p\in subtree(y),p!=y} [dep(p) \le dep(x)]

定义 g(x)=∑x∈subtree(p),p!=xf(x,p)g(x) = \sum_{x\in subtree(p),p!=x} f (x, p)

对于所有 i∈[1,n]i \in [1, n],求 g(i)g(i)

n≤5∗105n\le 5*10^5​

CF860E

Solution

考虑每个pp对g(i)g(i)的贡献，不难发现g(i)=∑dep(p)≤dep(i),p≠idep(lca(p,i))g(i) = \sum_{dep(p) \le dep(i),p\ne i} dep(lca(p,i))

考虑把所有点按深度从小到大排序，相同深度的同时加入，每加入一个点就把它到根这条链上的节点权值+1

每个点的答案就是它到根链上的权值和

树剖维护，时间复杂度O(nlog2n)O(n \log^2 n)，CF上能跑过

这道题还有O(n)O(n)长链剖分的高妙做法，但是并不会

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 = 5e5 + 100;

int N, fa[Maxn], root, dep[Maxn], dfn[Maxn], dfs_clock, size[Maxn], son[Maxn], top[Maxn];
int e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1];

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

vector <int> vec[Maxn];
int Maxdep;

inline void dfs (int x)
{
	dep[x] = dep[fa[x]] + 1, size[x] = 1;
	Chkmax(Maxdep, dep[x]);
	vec[dep[x]].pb(x);
	for (int i = Begin[x]; i; i = Next[i]) 
	{ 
		int y = To[i]; 
		dfs(y); 
		size[x] += size[y];
		if (size[y] > size[son[x]]) son[x] = y;
	}
}

inline void dfs (int x, int now)
{
	dfn[x] = ++dfs_clock, top[x] = now;
	if (son[x]) dfs(son[x], now);
	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (y == fa[x] || y == son[x]) continue;
		dfs(y, y);
	}
}

LL Ans[Maxn];

namespace SEG
{
#define ls (root << 1)
#define rs (root << 1 | 1)
#define lson ls, l, mid
#define rson rs, mid + 1, r
	struct tree
	{
		LL sum, tag;
	} Tree[Maxn << 2];

	inline void push_up (int root) { Tree[root].sum = Tree[ls].sum + Tree[rs].sum; }
	inline void push_down (int root, int l, int r, int mid)
	{
		if (!Tree[root].tag) return ;
		Tree[ls].tag += Tree[root].tag, Tree[rs].tag += Tree[root].tag;
		Tree[ls].sum += Tree[root].tag * (mid - l + 1), Tree[rs].sum += Tree[root].tag * (r - mid);
		Tree[root].tag = 0;
	}

	inline void update (int root, int l, int r, int x, int y, int z)
	{
		if (x <= l && r <= y) { Tree[root].sum += 1ll * (r - l + 1) * z; Tree[root].tag += z; return ; }
		int mid = l + r >> 1;
		push_down (root, l, r, mid);
		if (x <= mid) update (lson, x, y, z);
		if (y > mid) update (rson, x, y, z);
		push_up (root);
	}

	inline LL query (int root, int l, int r, int x, int y)
	{
		if (x <= l && r <= y) return Tree[root].sum;
		int mid = l + r >> 1; LL ans = 0;
		push_down(root, l, r, mid);
		if (x <= mid) ans += query (lson, x, y);
		if (y > mid) ans += query (rson, x, y);
		return ans;
	}

#undef ls
#undef rs
#undef lson
#undef rson
}

inline void Update (int x)
{
	while (x)
	{
		SEG :: update (1, 1, N, dfn[top[x]], dfn[x], 1);
		x = fa[top[x]];
	}
}

inline LL Query (int x)
{
	LL ans = 0;
	while (x)
	{
		ans += SEG :: query (1, 1, N, dfn[top[x]], dfn[x]);
		x = fa[top[x]];
	}
	return ans;
}

inline void Solve ()
{
	dfs(root);
	dfs(root, root);
	for (int i = 1; i <= Maxdep; ++i)
	{
		for (int j = 0; j < vec[i].size(); ++j) Update (vec[i][j]);
		for (int j = 0; j < vec[i].size(); ++j) Ans[vec[i][j]] = Query (vec[i][j]);
	}
	for (int i = 1; i <= N; ++i) printf("%lld ", Ans[i] - dep[i]);
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) fa[i] = read<int>(), add_edge (fa[i], i), (!fa[i]) ? (root = i) : 0;
}

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