Little Pony and Permutation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 846 Accepted Submission(s): 444
Problem Description
As a unicorn, the ability of using magic is the distinguishing feature among other kind of pony. Being familiar with composition and decomposition is the fundamental course for a young unicorn. Twilight Sparkle is interested in the decomposition of permutations. A permutation of a set S = {1, 2, ..., n} is a bijection from S to itself. In the great magician —— Cauchy's two-line notation, one lists the elements of set S in the first row, and then for each element, writes its image under the permutation below it in the second row. For instance, a permutation of set {1, 2, 3, 4, 5} σ can be written as:
Here σ(1) = 2, σ(2) = 5, σ(3) = 4, σ(4) = 3, and σ(5) = 1. Twilight Sparkle is going to decompose the permutation into some disjoint cycles. For instance, the above permutation can be rewritten as:
Help Twilight Sparkle find the lexicographic smallest solution. (Only considering numbers).
Input
Input contains multiple test cases (less than 10). For each test case, the first line contains one number n (1<=n<=10^5). The second line contains n numbers which the i-th of them(start from 1) is σ(i).
Output
For each case, output the corresponding result.
Sample Input
5
2 5 4 3 1
3
1 2 3
Sample Output
(1 2 5)(3 4)
(1)(2)(3)
题意：
题意较难理解，其实就是一个置换群！
看代码就会理解题意！
#include <iostream>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define inf 0x3f3f3f3f
#define debug(x) cout<<"---"<<x<<"---"<<endl
typedef long long ll;
using namespace std;
const int N = 100005;
int n, a[100010], vis[100010];
int main()
{
while (scanf("%d", &n) != EOF)
{
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
}
for (int i = 1; i <= n; i++)
{
if (vis[i] == 1)
{
continue;
}
int gg = i;
printf("(%d", gg);
vis[gg] = 1;
gg = a[gg];
while (vis[gg] == 0)
{
vis[gg] = 1;
printf(" %d", gg);
gg = a[gg];
}
printf(")");
}
printf("\n");
}
return 0;
}
