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

52 5 4 3 131 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 0x3f3f3f3fint 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;}