10-排序6 Sort with Swap(0, i) (25 分)

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 (≤10​^5​​) 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<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;

int main()
{
    int n;
    int num;    //数组存储方式调换，便于交换arr[0]和arr[arr[0]] 
    int arr[maxn] = {0};
    scanf("%d",&n);
    
    int left = n - 1; //控制循环次数，需要交换数的个数，0除外 
    for (int i = 0; i < n; i++)
    {
        scanf("%d",&num);
        arr[num] = i;
        if (num == i && num != 0)
        {
            left--;
        }
    }
    
    int k = 1;    //如果排序还未最终完成但是 0 已经回到arr[0]处，交换arr[k]值
    int ans = 0;    //纪录交换的次数 
    while (left > 0)
    {
        if (arr[0] == 0)
        {
            while (k < n)
            {
                if (arr[k] != k)
                {
                    swap(arr[0], arr[k]);
                    ans++;
                    break;
                }
                k++;
            }
        }
        else
        {
            swap(arr[0], arr[arr[0]]);
            ans++;
            left--;
        }
    }
    
    printf("%d", ans);
    return 0;
}