HDU 4985-Little Pony and Permutation（模拟置换）

Little Pony and Permutation

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 37   Accepted Submission(s) : 14

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)

 

Source

BestCoder Round #7

题目意思：

有1~N，给出这N个数的置换，判断哪些数是同一范围内的，用括号将同一范围内的数括起来。

解题思路：

用a[i]表示数i的置换数，比如题目给的2 5 4 3 1，表示a[1]=2,a[2]=5,a[3]=4,a[4]=3,a[5]=1.
因为要求字典序的解，所以我们从a[1]开始，每次找到a[i]表示的数然后标记其是否访问过，

例如a[1]=2,a[2]=5,a[5]=1，此时a[1]=2已经访问，所以这1,2,5三个数是同一个范围内的。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 100010
#define INF 0x3f3f3f3f
int a[MAXN];
bool vis[MAXN];
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("G:/cbx/read.txt","r",stdin);
//freopen("G:/cbx/out.txt","w",stdout);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
int n;
while(cin>>n)
{
memset(vis,false,sizeof(vis));
for(int i=1; i<=n; ++i)
cin>>a[i];
for(int i=1; i<=n; ++i)
{
bool s=true,e=true;//前后括号标记
int cnt=0;
if(vis[i]) continue;
while(1)
{
if(!vis[i])//若当前未访问
{
if(s)//加前括号
{
cout<<"("<<i;
s=false;
vis[i]=true;
i=a[i];//取指向的下一个数
}
else
{
vis[i]=true;
cout<<" "<<i;
i=a[i];
}
}
if(vis[i]) ++cnt;
if(cnt==n)//当前括号内的全部访问完毕
{
if(e)
{
cout<<")";//加后括号
e=false;
}
break;
}
}
}
cout<<endl;
}
return 0;
}