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

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本文探讨了一种特殊的排序问题，即如何仅通过交换数组中位置0的元素与其他任意位置的元素来对数组进行排序，并提供了一个有效的解决方案及其实现代码。

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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:

注意点：当arr[0]的value=0时是特殊情况，需要找到当前数组中第一个未在与自己下标相等的位置的节点，注意：每次在查找的时候不要从0开始找到N，后一次的查找都可以基于前一次的查找，例如：若前一次查找到K，那么0~K的数应该都已经在自己应该在的位置上了，下次从K+1开始查找就OK！

参考AC代码：

#include <iostream>
#include <vector>
#include <stdio.h>
using namespace std;
vector<int> arr(100001);
int N;
int main()
{
    scanf("%d",&N);
    int countNum = 0;
    for(int i=0;i<N;i++){
        scanf("%d",&arr[i]);
    }
    int findIt = 1;
    int start = 0;
    while(findIt){
        while(arr[0]!=0){
            swap(arr[0],arr[arr[0]]);
            countNum++;
        }
        findIt = 0;
        for(int j=start;j<N;j++){
            if(arr[j]!=j){
                findIt = j;
                start = j;
                swap(arr[0],arr[j]);
                countNum++;
                break;
            }
        }
    }
    printf("%d",countNum);
    return 0;
}