Pat1067代码

题目描述：

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:

AC代码：

#include
  
   
#define N 100005

using namespace std;

int Check(int A[],int begin,int end)//check the position of the data
{
	int i;
	for(i=begin;i<=end;i++)
		if(A[i]!=i)
			return i;
	return 0;
}

int main(int argc,char *argv[])
{
	int i,n;
	int Array[N];
	int ans=0,temp;
	int index;
	scanf("%d",&n);
	for(i=0;i
   
    
 解题思路： 分两种情况：1）A[0]!=0，则需要将元素0与所在位置元素进行交换，例如（4，0，2，1）那么0与1交换变成（4，1，2，0），如此循环下去，可是这样并不好操作每次都需要查找元素的位置进行交换，于是换一种思路：直接与A[0]进行交换，其结果是一样的。 2）A[0]==0,则只需将元素0与任何一个位置不正确的元素进行交换即可，然后重复第一种情况的操作。

Pat(Advanced Level)Practice--1067(Sort with Swap(0,*))

Pat1067代码