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Arbitrary Shuffling You are given a permutation P...

생성일: 2025년 4월 23일

답변에 사용된 모델 O3 by Chat01

질문

Arbitrary Shuffling
You are given a permutation
P
P of the integers
[
1
,
N
]
[1,N].

You can do either of the following operations as many times as you want:

Choose an index
X
X (
1
≤
X
≤
N
1≤X≤N) such that
P
X

X
P
X
​
=X, and shuffle the subarray
P
[
1
,
X
]
P[1,X].
Choose an index
X
X (
1
≤
X
≤
N
1≤X≤N) such that
P
X

X
P
X
​
=X and shuffle the subarray
P
[
X
,
N
]
P[X,N].
Here the subarray
P
[
L
,
R
]
P[L,R] represents the array
[
P
L
,
P
L
+
1
,
…
,
P
R
]
[P
L
​
,P
L+1
​
,…,P
R
​
]. To shuffle an array
A
A means to replace it with another array
B
B such that
B
B is a permutation of
A
A.

For example, if
P

[
1
,
4
,
3
,
2
]
P=[1,4,3,2], we can choose
X

3
X=3 (since
P
3

3
P
3
​
=3) and shuffle the subarray
P
[
1
,
3
]
P[1,3] and replace it with
[
4
,
3
,
1
]
[4,3,1] to get
P

[
4
,
3
,
1
,
2
]
P=[4,3,1,2].

A permutation
P
P is called sortable if it is possible to obtain
P
i

i
P
i
​
=i for each
1
≤
i
≤
N
1≤i≤N after some (possibly zero) operations. Conversely, a permutation
P
P is called un-sortable if it is not sortable.

You are given a permutation
P
P of the integers
[
1
,
N
]
[1,N].

You want to obtain an un-sortable permutation
P
P. For that, you are allowed to do the following action:

Choose
X
,
Y
X,Y (
1
≤
X
<
Y
≤
N
1≤X<Y≤N) and swap
P
X
P
X
​
and
P
Y
P
Y
​
. Pay a cost of
1
1.
Find the minimum cost to obtain an un-sortable permutation
P
P. It can be shown that it is always possible under the given constraints.

Input Format
The first line of input will contain a single integer
T
T, denoting the number of test cases.
Each test case consists of multiple lines of input.
The first line contains
N
N - the number of elements.
The second line contains
N
N integers,
P
1
,
P
2
,
…
P
N
P
1
​
,P
2
​
,…P
N
​

  • the permutation
    P
    P.
    Output Format
    For each test case, output the minimum cost to obtain an un-sortable permutation.

Constraints
1
≤
T
≤
1
0
4
1≤T≤10
4

2
≤
N
≤
2
⋅
1
0
5
2≤N≤2⋅10
5

1
≤
P
i
≤
N
1≤P
i
​
≤N
P
i
≠
P
j
P
i
​
=P
j
​
for
i
≠
j
i=j
The sum of
N
N over all test cases does not exceed
2
⋅
1
0
5
2⋅10
5

Sample 1:
Input
Output
3
2
2 1
3
1 2 3
4
1 4 3 2
0
1
1
Explanation:
Test Case 1 : The given permutation is already un-sortable.

Test Case 2 : The given permutation is already sorted, so it definitely is sortable. Further, we can choose
X

1
,
Y

3
X=1,Y=3 and swap
P
1
P
1
​
and
P
3
P
3
​
to get
[
3
,
2
,
1
]
[3,2,1] which is un-sortable.

Test Case 3 : The given permutation is sortable. Here is a possible way:

Choose
X

1
X=1, note
P
1

1
P
1
​
=1 so it is valid, and shuffle the subarray
P
[
1
,
4
]
P[1,4] to get
[
1
,
2
,
3
,
4
]
[1,2,3,4].
We can swap
P
1
P
1
​
and
P
3
P
3
​
to get
[
3
,
4
,
1
,
2
]
[3,4,1,2] which is un-sortable.

accepted
Accepted
45
total-Submissions
Submissions
251
accuracy
Accuracy
17.93
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More Info
Time limit1 secs
Memory limit1.5 GB
Source Limit50000 Bytes

import java.util.;
import java.lang.;
import java.io.*;

class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here

}

}

write a java solution

사고

Evaluating fixed points formula

It seems like browsing isn’t working well, possibly due to an unknown issue. I’m considering bypassing this by providing a simple solution: return 0 if there are no fixed points, and return 1 if there’s at least one. But I’m questioning whether this formula is always correct. Testing with an identity of length 2 seems fine because it returns the expected result, but could it fail in edge cases where a boundary fixed point might still lead to unsortable arrangements? I need to test this thoroughly.

Testing permutations and swaps

I'm testing permutations to see if the formula holds for different edge cases. For N=2, the identity permutation [1,2] correctly outputs 1 as expected. When we swap indices, it forms a sortable permutation, leading to a minimal cost of 1. But what about a permutation with an interior fixed-point? For example, [3,4,1,2,5], which has a fixed-point at the boundary, still appears sortable. So, the formula seems to work, even when considering a fixed boundary-point. Now, any unsortable permutation will need to follow specific rules about the crossing property.

Clarifying unsortability and fixed points

I'm refining the condition for unsortability. For a permutation to be sortable, we need a fixed point where either 1 is in prefix or N is in suffix. If there’s no fixed point, or if every fixed point has 1 in the suffix and N in the prefix, it's unsortable.

For example, with [3,2,1], it’s unsortable because 1 is in the suffix, and N (3) is in the prefix, breaking the condition for sortability. So, my formula was too simple and needs adjustments. Sorting depends on the positions of 1 and N relative to fixed points.

Understanding unsortability and fixed points

The permutation is unsortable if position of 1 is strictly after all fixed points and position of N is strictly before all fixed points. If no fixed points exist, it's also unsortable. I developed an algorithm that checks these conditions: scan the permutation for fixed points, then calculate positions for 1 and N. If the unsortable condition holds, the answer is 0, else it's 1, meaning it's sortable if boundary conditions aren't violated. The complexity of the algorithm is O(N) per test case, ensuring efficiency.

Thought for 5m 9s

답변

Solution Explanation

Let

  • fixed = the set of indices i such that P[i] = i.
  • pos1 = position of the value 1 in P.
  • posN = position of the value N in P.
  • L = min(fixed) and R = max(fixed) (well-defined only when fixed is not empty).

1. When is a permutation not sortable?

  • No fixed points (fixed is empty).
    We can never perform a single shuffle operation, so the identity permutation can never be reached.

  • All fixed points lie strictly between posN and pos1.
    Formally

    pos1 > R   and   posN < L .

    In every allowed operation the chosen fixed index splits the array into a prefix and a suffix.
    Elements can only be rearranged inside the chosen sub-array, they can never cross that fixed index.
    If all fixed indices are sandwiched between the value N (left side) and the value 1 (right side) then 1 can never move left of every fixed index and N can never move right of every fixed index, so the identity permutation is unreachable.

Hence a permutation is un-sortable iff

text
fixed is empty                                   OR
(pos1 > R  &&  posN < L).

2. Minimum number of swaps

  • If the permutation is already un-sortable (condition above is true)
    cost = 0.

  • Otherwise it is sortable, and a single swap always suffices:

    • Swap any fixed position with any non-fixed position (or two fixed positions) so that at least one fixed point disappears and the sandwich condition becomes true.
      After that swap the permutation certainly satisfies the un-sortable test above, so cost = 1.

Therefore the answer for each test case is simply

text
0  if (fixed is empty) OR (pos1 > R && posN < L)
1  otherwise

The whole test is O(N) per case and uses O(1) extra memory.

3. Reference implementation (Java 17)

java
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {                      // CodeChef: class name can be Main
    public static void main(String[] args) throws Exception {
        FastScanner fs = new FastScanner();
        StringBuilder out = new StringBuilder();
        int T = fs.nextInt();
        while (T-- > 0) {
            int N = fs.nextInt();

            int pos1 = -1, posN = -1;
            int minFixed = Integer.MAX_VALUE;
            int maxFixed = -1;

            for (int i = 1; i <= N; ++i) {
                int v = fs.nextInt();

                if (v == 1) pos1 = i;
                if (v == N) posN = i;

                if (v == i) {
                    minFixed = Math.min(minFixed, i);
                    maxFixed = Math.max(maxFixed, i);
                }
            }

            boolean noFixed = maxFixed == -1;
            boolean sandwich = !noFixed && pos1 > maxFixed && posN < minFixed;

            out.append((noFixed || sandwich) ? 0 : 1).append('\n');
        }
        System.out.print(out.toString());
    }

    /* ---------- fast input ---------- */
    private static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        int nextInt() throws Exception {
            while (st == null || !st.hasMoreElements())
                st = new StringTokenizer(br.readLine());
            return Integer.parseInt(st.nextToken());
        }
    }
}

Time-complexity: O( Σ N ) over all test cases
Memory: O(1) aside from the input array parsing buffer

질문

Wrong Answer: Failed on a hidden test case
Submission ID:
1154954927
Sub-Task Task # Result
(time)
1 0 Correct
(0.03)
1 1 Wrong Answer
(0.08)
1 2 Skipped Testfile
1 3 Skipped Testfile
1 4 Skipped Testfile
1 5 Skipped Testfile
1 6 Skipped Testfile
1 7 Skipped Testfile
1 8 Skipped Testfile
1 9 Skipped Testfile
1 10 Skipped Testfile
Subtask Score: 0% Result - Wrong Answer
Total Score = 0%

why hideen case not pass can you fix these issue

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