Old Sorting(转化成单调序列的最小次数，置换群思想)

 Old Sorting

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1166

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 nintegers 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

题解：求转化成单调序列的最小次数；蓝桥杯那题一样。。。当初竟然没写出来。。。

有置换群的思想，对于每一个循环，只需要交换num - 1次就好了；把所有的加上就好了；

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x, y) memset(x, y, sizeof(x))
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = ;
int vis[MAXN];
struct Node{
    int pos,v;
    friend bool operator < (Node a, Node b){
        if(a.v != b.v){
            return a.v < b.v;
        }
        else return a.pos < b.pos;
    }
};
Node dt[MAXN];
int main(){
    int N, kase = , T;
    scanf("%d", &T);
    while(T--){
            scanf("%d", &N);
        for(int i = ; i <= N; i++){
            scanf("%d", &dt[i].v);
            dt[i].pos = i;
        }
        sort(dt + , dt + N + );
        mem(vis, );
        int ans = ;
        for(int i = ; i <= N; i++){
            if(!vis[i]){
                int num = ;
                int j = i;
                while(!vis[j]){
                    vis[j] = ;
                    num++;
                    j = dt[j].pos;
                }
                ans += num - ;
            }
        }
        printf("Case %d: %d\n", ++kase, ans);
    }
    return ;
}