codeforces#P1787F. Inverse Transformation
Inverse Transformation
Description
A permutation scientist is studying a self-transforming permutation $a$ consisting of $n$ elements $a_1,a_2,\ldots,a_n$.
A permutation is a sequence of integers from $1$ to $n$ of length $n$ containing each number exactly once. For example, $[1]$, $[4, 3, 5, 1, 2]$ are permutations, while $[1, 1]$, $[4, 3, 1]$ are not.
The permutation transforms day by day. On each day, each element $x$ becomes $a_x$, that is, $a_x$ becomes $a_{a_x}$. Specifically:
- on the first day, the permutation becomes $b$, where $b_x = a_{a_x}$;
- on the second day, the permutation becomes $c$, where $c_x = b_{b_x}$;
- $\ldots$
You're given the permutation $a'$ on the $k$-th day.
Define $\sigma(x) = a_x$, and define $f(x)$ as the minimal positive integer $m$ such that $\sigma^m(x) = x$, where $\sigma^m(x)$ denotes $\underbrace{\sigma(\sigma(\ldots \sigma}_{m \text{ times}}(x) \ldots))$.
For example, if $a = [2,3,1]$, then $\sigma(1) = 2$, $\sigma^2(1) = \sigma(\sigma(1)) = \sigma(2) = 3$, $\sigma^3(1) = \sigma(\sigma(\sigma(1))) = \sigma(3) = 1$, so $f(1) = 3$. And if $a=[4,2,1,3]$, $\sigma(2) = 2$ so $f(2) = 1$; $\sigma(3) = 1$, $\sigma^2(3) = 4$, $\sigma^3(3) = 3$ so $f(3) = 3$.
Find the initial permutation $a$ such that $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$ is minimum possible.
Each test contains multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$; $1 \le k \le 10^9$) — the length of $a$, and the last day.
The second line contains $n$ integers $a'_1,a'_2,\ldots,a'_n$ ($1 \le a'_i \le n$) — the permutation on the $k$-th day.
It's guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$.
For each test case, if at least one initial $a$ consistent with the given $a'$ exists, print "YES", then print $n$ integers $a_1,a_2,\ldots,a_n$ — the initial permutation with the smallest sum $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$. If there are multiple answers with the smallest sum, print any.
If there are no valid permutations, print "NO".
Input
Each test contains multiple test cases. The first line contains an integer $t$ ($1 \le t \le 10^4$) — the number of test cases.
The first line of each test case contains two integers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$; $1 \le k \le 10^9$) — the length of $a$, and the last day.
The second line contains $n$ integers $a'_1,a'_2,\ldots,a'_n$ ($1 \le a'_i \le n$) — the permutation on the $k$-th day.
It's guaranteed that the sum of $n$ does not exceed $2 \cdot 10^5$.
Output
For each test case, if at least one initial $a$ consistent with the given $a'$ exists, print "YES", then print $n$ integers $a_1,a_2,\ldots,a_n$ — the initial permutation with the smallest sum $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$. If there are multiple answers with the smallest sum, print any.
If there are no valid permutations, print "NO".
10
5 3
1 2 3 4 5
7 2
1 2 3 4 5 6 7
8 998
1 2 3 4 5 6 7 8
6 1
6 3 5 4 1 2
4 8
4 2 1 3
9 1
1 5 4 8 7 6 3 2 9
5 9999999
2 3 4 5 1
7 97843220
4 6 1 2 7 5 3
3 1000000000
2 1 3
12 3
8 9 10 1 5 3 11 4 7 6 12 2
YES
2 3 4 1 5
YES
6 2 5 7 1 3 4
YES
2 3 4 5 6 7 8 1
YES
3 1 6 4 2 5
YES
4 2 1 3
NO
YES
3 4 5 1 2
YES
2 5 4 6 3 7 1
NO
YES
3 7 8 6 5 1 12 10 11 4 2 9
Note
In the second test case, the initial permutation can be $a = [6,2,5,7,1,3,4]$, which becomes $[3,2,1,4,6,5,7]$ on the first day and $a' = [1,2,3,4,5,6,7]$ on the second day (the $k$-th day). Also, among all the permutations satisfying that, it has the minimum $\sum\limits^n_{i=1} \dfrac{1}{f(i)}$, which is $\dfrac{1}{4}+\dfrac{1}{1}+\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{2}=3$.
In the fifth test case, the initial permutation can be $a = [4,2,1,3]$, which becomes $[3,2,4,1]$ on the first day, $[4,2,1,3]$ on the second day, and so on. So it finally becomes $a' = [4,2,1,3]$ on the $8$-th day (the $k$-th day). And it has the minimum $\sum\limits^n_{i=1} \dfrac{1}{f(i)} = \dfrac{1}{3} + \dfrac{1}{1} + \dfrac{1}{3} + \dfrac{1}{3} = 2$.