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 = 10010; 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 = 0, T; scanf("%d", &T); while(T--){ scanf("%d", &N); for(int i = 1; i <= N; i++){ scanf("%d", &dt[i].v); dt[i].pos = i; } sort(dt + 1, dt + N + 1); mem(vis, 0); int ans = 0; for(int i = 1; i <= N; i++){ if(!vis[i]){ int num = 0; int j = i; while(!vis[j]){ vis[j] = 1; num++; j = dt[j].pos; } ans += num - 1; } } printf("Case %d: %d\n", ++kase, ans); } return 0; }
Old Sorting(转化成单调序列的最小次数，置换群思想)
原文：http://www.cnblogs.com/handsomecui/p/5413999.html