1067. Sort with Swap(0,*)

最新推荐文章于 2021-04-25 19:32:33 发布

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最新推荐文章于 2021-04-25 19:32:33 发布

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分类专栏： PAT 文章标签： PAT C++

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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。文章提供了一个算法解决方案，并分析了不同情况下所需的最小交换次数。

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

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

如果有k个数形成一个环，如果0也在这个环里，那么需要交换k-1次，否则就需要k+1次,即任意一元素与0交换形成k+1个元素的环。

记录环的个数和不在自己位置的数的个数。由于初始只能有<=1个环包含0，分一下情况。

把交换的过程换成直接赋值，会省一些时间吧~~~

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int main()
{
	int n;
	cin>>n;
	vector<int>num(n);
	for(int i=0;i<n;++i)
		cin>>num[i];
	int cnt=0,cntn=0;
	bool flag=(num[0]==0);
	for(int i=0;i<n;++i)
	{
		int s=i;
		if(num[s]!=s)
			cnt++;
		while(num[s]!=s)
		{
			int next=num[s];
			num[s]=s;
			cntn++;
			s=next;
		}
	}
	if(flag)
		cout<<cnt+cntn<<endl;
	else
		cout<<max(0,cntn+cnt-2)<<endl;
	return 0;
}