hdu 4985(模拟)

Little Pony and Permutation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 639    Accepted Submission(s): 342

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)

 

= = 把a数组设为bool 型，无限WA。。。我真是。。。本来1A的

题意:就是找循环。比如说 1 ->2 -> 5->1 所以 (1,2,5)是一个循环。

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;
const int N = 100005;
bool vis[N];
int res[N],a[N];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++){
            int j=i;
            if(vis[j]) continue;
            int id = 0;
            while(!vis[j]){
                res[id++] = j;
                vis[j]=true;
                j = a[j];
            }
            printf("(");
            for(int i=0;i<id-1;i++){
                printf("%d ",res[i]);
            }
            printf("%d)",res[id-1]);
        }
        printf("
");
    }
    return 0;
}