A1067 Sort with Swap(0, i) (25 point(s))

Swap(0, i)

1. 原文

原题

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

2. 解析

交换swap(0,i)

• 先确定a[0]=0; while(a[0]!=0)
• 再交换swap(0,i) if (a[i]!=i)

3. AC代码

#include<cstdio>
#include<algorithm>
using namespace std;
int a[100010]={};
int main()
{
	int n;
	scanf("%d",&n);
	int num;
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&num);
		a[num]=i;
	}
	int ans=0;
	for (int i = 0; i < n; ++i)
	{
		if (a[i]!=i)
		{
			while(a[0]!=0){
				swap(a[0],a[a[0]]);
				ans++;
			}
			if (a[i]!=i)
			{
				swap(a[0],a[i]);
				ans++;
			}
		}
	}
	printf("%d\n", ans);
	return 0;
}