10-排序6 Sort with Swap(0, i) 分数 25

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输入一个包含非负整数的数组，计算使用Swap(0,*)操作进行排序所需的最少交换次数。算法涉及读入数组、记录反函数并查找0的位置进行交换。

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:

思路：

读入数组同时记录其反函数，即值为i的数下标为h[i]

当前a[0]==0时，进行检查，查找数组中是否有a[i]!=i，并记录当前检查到第几个数（这样时间复杂度是O（n）,不然会超时，可自行尝试）

然后进行a[i]与a[0]的交换。

当前a[0]!=0时，找到0的下标（h[0]），让数h[0]与数0进行交换

#include <bits/stdc++.h>
using namespace std;
int n,k,m,cnt = 0;
int a[100005];
int h[100005];
int checkk = 0;
bool check(){
	for(int i = checkk;i<n;i++){
		if(a[i] != i){
			checkk = i;
			a[0] = a[i];
			a[i] = 0;
			h[0] = i;
			h[a[0]] = 0;
			cnt++;
			return false;
		}
	}
	return true;
}
int main()
{
	cin>>n;
	for(int i = 0;i<n;i++){
		cin>>a[i];
		h[a[i]] = i;
	}
	while(1){
		if(a[0] == 0){
			if(check())break;
		}else {
			int d = h[0];
			int dd = h[d];
			a[d] = d;
			a[dd] = 0;
			h[a[d]] = d;
			h[0] = dd;
			cnt++;
		}
	}
	cout<<cnt;
	return 0;
}