PAT A1067. Sort with Swap(0,*) (25)_a1067 sort with-优快云博客

本文探讨了一种特殊的排序问题——仅允许使用Swap(0, *)操作来对0到N-1的排列进行排序。通过具体实例说明了如何实现，并提供了一段C++代码来计算最少交换次数。

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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:

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
#include <string>
#define Max 100010
using namespace std;
int main()
{
	int n,S[Max];
	scanf("%d",&n);
	int f=n-1;
	int m=0,l;
	for(int i=0;i<n;i++)
	{
		scanf("%d",&S[i]);
		if(S[i]==i) f--;
	}
	while(f>0)
	{
		if(S[0]==0)   //0在本位
		{
			int k=1;
			while(k<n)
			{
				if(S[k]!=k)
				{
					swap(S[0],S[k]);
					m++;
					break;
				}
				k++;
			}
		}
		else if(S[0]!=0)
		{
			swap(S[0],S[S[0]]);
		    m++;
			f--;
		}
	}
	printf("%d\n",m);
    system("pause");
	return 0;
}