学步园

Given a permutation of numbers from 1 to n, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence is 2, 3, 5, 4, 1, we can sort them in the following way:

2 3 5 4 1
1 3 5 4 2
1 3 2 4 5
1 2 3 4 5

Here three swaps have been used. The problem is, given a specific permutation, how many swaps we needs to take at least.

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each case contains two lines. The first line contains the integer n (1 <= n <= 10000), and the second line gives the initial permutation.

For each test case, the output will be only one integer, which is the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation.

2
3
1 2 3
5
2 3 5 4 1

0
3

解析：

很明显 2 3 5 1构成一个环 4 构成一个环

为了保证最少的次数，只要在每个环上做变化

假设r个环，每个环上的个数是h[i] ，则每个环需要swap h[i]-1次就可以

则总的次数是 (h[1]-1)+...+(h[r]-1)=(h[1]+...+h[r])-r 即 答案=总的数-环的个数