lightoj 1166 - Old Sorting 【置换群】

题目链接：lightoj 1166 - Old Sorting

1166 - Old Sorting

PDF (English) Statistics Forum

Time Limit: 2 second(s) Memory Limit: 32 MB

Given an array containing a permutation of 1 to n, you have to find the minimum number of swaps to sort the array in ascending order. A swap means, you can exchange any two elements of the array.

For example, let n = 4, and the array be 4 2 3 1, then you can sort it in ascending order in just 1 swaps (by swapping 4 and 1).

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains two lines, the first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers separated by spaces. You may assume that the array will always contain a permutation of 1 to n.

Output

For each case, print the case number and the minimum number of swaps required to sort the array in ascending order.

Sample Input

Output for Sample Input

3

4

4 2 3 1

4

4 3 2 1

4

1 2 3 4

Case 1: 1

Case 2: 2

Case 3: 0

题意：给定n的一个排列，问你最少交换多少次才能使序列升序。

思路：一个元素个数为cnt的置换群最少需要cnt-1交换。

AC代码：

#include <cstdio>
bool vis[110];
int a[110], pos[110];
int main()
{
int t, kcase = 1;
scanf("%d", &t);
while(t--) {
int n; scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
pos[a[i]] = i; vis[i] = false;
}
int ans = 0;
for(int i = 1; i <= n; i++) {
if(vis[a[i]]) continue;
vis[a[i]] = true;
int cnt = 1; int next = a[i];
while(pos[next] != a[i]) {
next = pos[next]; cnt++;
vis[next] = true;
}
//printf("%d\n", cnt);
ans += cnt - 1;
}
printf("Case %d: %d\n", kcase++, ans);
}
return 0;
}