【PAT（甲级）】1067 Sort with Swap(0, i)（附解题思路）

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

解题思路：

题目给出一组已经打乱排序的数组，让你只能交换值为0和其他的数字，使得最终排序为升序，要求输出最少的交换次数。

1. 首先我们要明确，值为0的数，无论怎么交换，最终都会出现在数组下标为0的位置上。所以我们第一步就是从数组下标为0的位置开始交换，因为是两两交换，每次都会使得至少一个位置上的数字正确，所以肯定能把0放到下标为0的位置上：

while(number[0]!=0){
                swap(number[0],number[number[0]]);
                times++;
            } // 使得number[0]位置上的数字可以出现在正确的位置上

2. 如果此时，依旧没有使得number[i]位置上的数字正确，那么就将0放到number[i]上，在第二次使0归为的途中，number[i]就一定会出现一个正确的数字了。

代码：

#include
using namespace std;
 
int main(){
	int N;
	cin>>N;
	int number[N];
	int times = 0;
	for(int i=0;i
		cin>>number[i];
	}
	
	for(int i=1;i
		if(number[i] != i){
            while(number[0]!=0){
                swap(number[0],number[number[0]]);
                times++;
            }
            if(number[i]!=i){
                swap(number[0],number[i]);
                times++;
            }
        }
		
	}
	cout<
}