PAT 1067. Sort with Swap(0,*) (25)

1067. Sort with Swap(0,*) (25)

Given any permutation of the numbers {0, 1, 2,..., N-1}, it is easy to sort them in increasing order. But what if Swap(0, *) is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (<=10⁵) followed by a permutation sequence of {0, 1, ..., N-1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10 3 5 7 2 6 4 9 0 8 1

Sample Output:

9

#include <iostream>
#include <set>

using namespace std;

set<int> WrongIndex;
int sequence[100000];

int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        int num;
        cin >> num;
        sequence[i] = num;
        if (num != i)
            WrongIndex.insert(i);
    }
    int totalNum = WrongIndex.size();
    int cycle = 0;
    while (!WrongIndex.empty())
    {
        int initial = *(WrongIndex.begin());
        int next = sequence[initial];
        WrongIndex.erase(initial);
        while (next != initial)
        {
            WrongIndex.erase(next);
            next = sequence[next];
        }
        cycle++;
    }
    if (totalNum == 0)
        cout << 0;
    else if (sequence[0] != 0)
        cout << totalNum + cycle - 2;
    else
        cout << totalNum + cycle;
}