Description：

Kayaking is playing a puzzle game containing n different blocks. He marks the blocks with integers from 1 to n, which show the blocks’ original positions. Each time he can exchange two blocks and he wants to know how many times he needs at least to restore the puzzle.

Input

The input starts with one line contains exactly one positive integer which is the number of test cases. 
Each test case contains two lines. 
The first line contains an integer, which indicates the number of puzzle pieces. 
The second line contains n different integers, the i-th number means the mark of the block in the i-th position.

Output

For each test case, output one line with one number represents the minimum operations.

Sample Input

2
4
2 3 4 1
4
2 1 4 3

Sample Output

3
2

 

Hint

1<=T<=20,1<=n<=100000

 题意：

   

一开始以为是模拟冒泡排序结果会超时。

然后优化到一重循环。

AC代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define maxn 100005
using namespace std;
char str[maxn];
typedef long long ll;
int a[maxn],id[maxn];
int main()
{
    int t;
    while (~scanf("%d",&t))
    {
        while (t--)
        {
            int n;
            scanf("%d",&n);
            for (int i=1; i<=n; i++)
            {
                scanf("%d",&a[i]);
                id[a[i]]=i;
            }
            int ans=0;
            for (int i=1; i<=n; i++)
            {
                if (a[i]!=i)
                {
                    ans++;
                    int tmp=a[i],idd=id[i];
                    swap(a[i],a[id[i]]);
                    swap(id[tmp],id[i]);
                }
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}