PTA(Advanced Level)1067.Sort with Swap(0, i)

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:

9

思路

• 贪心策略：
  • 最好的置换结果就是一次就把数字还原到最后对应的位置上去
  • 可能出现的结果是0在不断置换的过程中回到了0的位置，但是整个数组仍非有序的，此时0要是和已经归位的数字发生替换那么就会多出步数，此时应该和还没归位的数字换位置

代码

#include<bits/stdc++.h>
using namespace std;
int a[100010];
int main()
{
	int n;
	cin >> n;
	for(int i=0;i<n;i++)	cin >> a[i];
	int remain = 0;
	for(int i=0;i<n;i++)
		if(a[i] != i && a[i] != 0)
			remain++;
	int k = 1;    //表示下一个0该跳转的位置(如果0跳到0上)
	int ans = 0;
	while(remain > 0)
	{
		if(a[0] == 0)    //0在本位上
		{
			while(k < n)
			{
				if(a[k] != k)   //找到第一个不在该在的位置上的数字
				{
					swap(a[0], a[k]);
					ans++;
					break;
				}
				k++;
			}
		}

		while(a[0] != 0)    //当0不在0号位的时候
		{
			swap(a[0], a[a[0]]);   //将0所在位置上的数和0进行交换
			ans++;
			remain--;
		}
	}
	cout << ans;
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805403651522560