1067. Sort with Swap(0,*)

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 <string>#include <map>#include <algorithm>#include <vector>using namespace std;int main(){  int n;  cin >> n;  int* arr = new int[n];  bool* flag = new bool[n];  int count(0);  int circlecount(0);  for(int i = 0;i < n;++ i)  {    cin >> arr[i];    flag[i] = false;  }  for(int i = 0;i < n;++ i)  {    if(flag[i] == true)      continue;    if(arr[i] == i)    {      flag[i] = true;    }    else    {      ++ circlecount;      int idx = i;      int value = arr[idx];      while(value != i)      {        idx = value;        value = arr[idx];        flag[idx] = true;        ++ count;      }    }  }  if(arr[0] != 0)    cout << count + 2 * (circlecount - 1);  else    cout << count + 2 * circlecount;}