1067. Sort with Swap(0,*)

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:

9

#include<stdio.h>
#include<vector>
using namespace std;

int main()
{
	//freopen("F://Temp/input.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	
	vector<int> v;
	for(int i = 0; i < n; i ++)
	{
		int x;
		scanf("%d", &x);
		v.push_back(x);
	}
	
	int count  = 0;
	bool flag = 0;
	//0是不是在正确位置上 
	if(v[0] == 0)
		flag = true;
	int num = 0;//组数目
	
	for(int i = 0; i < n; i ++)
	{
		bool newGroup = false;
		while(v[i] != i)
		{
			newGroup = true;
			int tmp = v[v[i]];
			v[v[i]] = v[i];
			v[i] = tmp;
			count ++;
		}
		if(newGroup)
			num ++;
	}
	
	if(flag)
		count += num*2;
	else
		count += (num-1)*2;	
	
	printf("%d\n", count);
	return 0;
}