LightOJ - 1166 Old Sorting （置换群）

LightOJ - 1166

Old Sorting

Time Limit: 2000MSMemory Limit: 32768KB64bit IO Format: %lld & %llu

Submit Status

Description

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

3

4

4 2 3 1

4

4 3 2 1

4

1 2 3 4

Sample Output

Case 1: 1

Case 2: 2

Case 3: 0

Source

Problem Setter: Jane Alam Jan

//题意：

给你n个数，和这n个数的序列，可以两两交换，问把它们变换为从小到大的序列，最少变换几次？

//思路：

就是一个简单的置换群，没啥说的

#include<stdio.h>#include<string.h>#include<algorithm>#define N 110using namespace std;struct zz{int x;int id;}p[N];bool cmp(zz a,zz b){return a.x<b.x;}int a[N],vis[N];int main(){int t,T=1,n,i,j,k;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);p[i].x=a[i];p[i].id=i;}sort(p+1,p+n+1,cmp);int cnt,sum=0;memset(vis,0,sizeof(vis));for(i=1;i<=n;i++){cnt=0;k=p[i].x;if(!vis[k]){while(a[i]!=k){vis[k]=1;vis[a[i]]=1;k=p[k].id;cnt++;}sum+=cnt;}}printf("Case %d: %d\n",T++,sum);}return 0;}