Codeforces傳送門

洛谷傳送門

題目描述

Arkady words in a large company. There are nn employees working in a system of a strict hierarchy. Namely, each employee, with an exception of the CEO, has exactly one immediate manager. The CEO is a manager (through a chain of immediate managers) of all employees.

Each employee has an integer rank. The CEO has rank equal to $1$ , each other employee has rank equal to the rank of his immediate manager plus $1$

1 .

Arkady has a good post in the company, however, he feels that he is nobody in the company’s structure, and there are a lot of people who can replace him. He introduced the value of replaceability. Consider an employee $a$

a and an employee $b$ , the latter being manager of aa (not necessarily immediate). Then the replaceability $r(a,b)$ of $a$ with respect to $b$ is the number of subordinates (not necessarily immediate) of the manager $b$ , whose rank is not greater than the rank of $a$ . Apart from replaceability, Arkady introduced the value of negligibility. The negligibility $z_{a}$ of employee aa equals the sum of his replaceabilities with respect to all his managers, i.e. , where the sum is taken over all his managers $b$ .

Arkady is interested not only in negligibility of himself, but also in negligibility of all employees in the company. Find the negligibility of each employee for Arkady.

輸入輸出格式

輸入格式：

The first line contains single integer $n$ ( $1\le n\le 5·10^{5}$) — the number of employees in the company.

The second line contains $n$ integers $p_{1},p_{2},...,p_{n}$ ( $0\le p_{i}\le n$ ), where $p_{i}=0$ if the $i$ -th employee is the CEO, otherwise $p_{i}$ equals the id of the immediate manager of the employee with id $i$ . The employees are numbered from $1$ to $n$ . It is guaranteed that there is exactly one $0$ among these values, and also that the CEO is a manager (not necessarily immediate) for all the other employees.

輸出格式：

Print $n$ integers — the negligibilities of all employees in the order of their ids: $z_{1},z_{2},...,z_{n}$.

輸入輸出樣例

輸入樣例#1：

4
0 1 2 1

輸出樣例#1：

0 2 4 2 

輸入樣例#2：

5
2 3 4 5 0

輸出樣例#2：

10 6 3 1 0 

輸入樣例#3：

5
0 1 1 1 3

輸出樣例#3：

0 3 3 3 5 

說明

Consider the first example:

• The CEO has no managers, thus $z_{1}=0$ .
• $r(2,1)=2$ (employees $2$ and $4$ suit the conditions, employee $3$ has too large rank). Thus $z_{2}=r(2,1)=2$.
• Similarly,$ z_{4}=r(4,1)=2$ .
• $r(3,2)=1$ (employee $3$ is a subordinate of $2$ and has suitable rank). $r(3,1)=3$(employees $2$ , $3$ , $4$ suit the conditions). Thus $z_{3}=r(3,2)+r(3,1)=4$.

題目大意

給你一棵有根樹， 設 $b$為 $a$的祖先節點， $r(a,b)$表示在 $b$子樹中不超過 $a$深度的點的個數(不包括 $a$自己)， 求對於所有點的 $z(i)=\sum_{j\in ancestor(i)}r(i,j)$。

解題分析

很容易發現每個點 $a$對於其他點 $b$( $dep[a]\le dep[b]$)的貢獻就是 $dep[lca(a,b)]$, 所以將所有點按深度從小到大排序， 每次處理同一深度的點， 在其到根節點的路徑上區間 $+1$， 然後查詢其到根節點的路徑上的權值和即可， 總複雜度 $O(nlog^2(n))$， 卡卡常就過了。

這裡有一種 $O(nlog(n))$的做法。我們可以發現 $z(i)=z(fat[i])+dep[i]+\sum_{dep[i]=dep[j]}dep[lca(i,j)]$，那麼實際上我們只需要考慮同深度的點之間的貢獻。

將同深度的點按 $DFS$序排序， 那麼可以發現對於一個排在第 $i$位的點， 其前面的點從 $1$開始， 它們的 $lca$

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