Preparando MOJI
Santa has an infinite number of candies for each of $$$m$$$ flavours. You are given a rooted tree with $$$n$$$ vertices. The root of the tree is the vertex $$$1$$$. Each vertex contains exactly one candy. The $$$i$$$-th vertex has a candy of flavour $$$f_i$$$.
Sometimes Santa fears that candies of flavour $$$k$$$ have melted. He chooses any vertex $$$x$$$ randomly and sends the subtree of $$$x$$$ to the Bakers for a replacement. In a replacement, all the candies with flavour $$$k$$$ are replaced with a new candy of the same flavour. The candies which are not of flavour $$$k$$$ are left unchanged. After the replacement, the tree is restored.
The actual cost of replacing one candy of flavour $$$k$$$ is $$$c_k$$$ (given for each $$$k$$$). The Baker keeps the price fixed in order to make calculation simple. Every time when a subtree comes for a replacement, the Baker charges $$$C$$$, no matter which subtree it is and which flavour it is.
Suppose that for a given flavour $$$k$$$ the probability that Santa chooses a vertex for replacement is same for all the vertices. You need to find out the expected value of error in calculating the cost of replacement of flavour $$$k$$$. The error in calculating the cost is defined as follows.
$$$$$$ Error\ E(k) =\ (Actual Cost\ –\ Price\ charged\ by\ the\ Bakers) ^ 2.$$$$$$
Note that the actual cost is the cost of replacement of one candy of the flavour $$$k$$$ multiplied by the number of candies in the subtree.
Also, sometimes Santa may wish to replace a candy at vertex $$$x$$$ with a candy of some flavour from his pocket.
You need to handle two types of operations:
The first line of the input contains four integers $$$n$$$ ($$$2 \leqslant n \leqslant 5 \cdot 10^4$$$), $$$m$$$, $$$q$$$, $$$C$$$ ($$$1 \leqslant m, q \leqslant 5 \cdot 10^4$$$, $$$0 \leqslant C \leqslant 10^6$$$) — the number of nodes, total number of different flavours of candies, the number of queries and the price charged by the Bakers for replacement, respectively.
The second line contains $$$n$$$ integers $$$f_1, f_2, \dots, f_n$$$ ($$$1 \leqslant f_i \leqslant m$$$), where $$$f_i$$$ is the initial flavour of the candy in the $$$i$$$-th node.
The third line contains $$$n - 1$$$ integers $$$p_2, p_3, \dots, p_n$$$ ($$$1 \leqslant p_i \leqslant n$$$), where $$$p_i$$$ is the parent of the $$$i$$$-th node.
The next line contains $$$m$$$ integers $$$c_1, c_2, \dots c_m$$$ ($$$1 \leqslant c_i \leqslant 10^2$$$), where $$$c_i$$$ is the cost of replacing one candy of flavour $$$i$$$.
The next $$$q$$$ lines describe the queries. Each line starts with an integer $$$t$$$ ($$$1 \leqslant t \leqslant 2$$$) — the type of the query.
If $$$t = 1$$$, then the line describes a query of the first type. Two integers $$$x$$$ and $$$w$$$ follow ($$$1 \leqslant x \leqslant n$$$, $$$1 \leqslant w \leqslant m$$$), it means that Santa replaces the candy at vertex $$$x$$$ with flavour $$$w$$$.
Otherwise, if $$$t = 2$$$, the line describes a query of the second type and an integer $$$k$$$ ($$$1 \leqslant k \leqslant m$$$) follows, it means that you should print the expected value of the error in calculating the cost of replacement for a given flavour $$$k$$$.
The vertices are indexed from $$$1$$$ to $$$n$$$. Vertex $$$1$$$ is the root.
Output the answer to each query of the second type in a separate line.
Your answer is considered correct if its absolute or relative error does not exceed $$$10^{-6}$$$.
Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. The checker program considers your answer correct if and only if $$$\frac{|a-b|}{max(1,b)}\leqslant 10^{-6}$$$.
3 5 5 7
3 1 4
1 1
73 1 48 85 89
2 1
2 3
1 2 3
2 1
2 3
2920.333333333333
593.000000000000
49.000000000000
3217.000000000000
For $$$1$$$-st query, the error in calculating the cost of replacement for flavour $$$1$$$ if vertex $$$1$$$, $$$2$$$ or $$$3$$$ is chosen are $$$66^2$$$, $$$66^2$$$ and $$$(-7)^2$$$ respectively. Since the probability of choosing any vertex is same, therefore the expected value of error is $$$\frac{66^2+66^2+(-7)^2}{3}$$$.
Similarly, for $$$2$$$-nd query the expected value of error is $$$\frac{41^2+(-7)^2+(-7)^2}{3}$$$.
After $$$3$$$-rd query, the flavour at vertex $$$2$$$ changes from $$$1$$$ to $$$3$$$.
For $$$4$$$-th query, the expected value of error is $$$\frac{(-7)^2+(-7)^2+(-7)^2}{3}$$$.
Similarly, for $$$5$$$-th query, the expected value of error is $$$\frac{89^2+41^2+(-7)^2}{3}$$$.
Informação
Provedor Codeforces
Código CF960H
Tags
Submetido 0
BOUA! 0
Taxa de BOUA's 0%
Datas 09/05/2023 09:33:27
Relacionados