杭电4985 Little Pony and Permutation（水置换）

最新推荐文章于 2017-03-20 15:07:39 发布

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最新推荐文章于 2017-03-20 15:07:39 发布

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本文介绍了一个关于排列及其分解的问题，通过实例展示了如何将一个排列分解为若干不相交的循环，并给出了一个C++实现代码，旨在帮助理解排列的基础概念及其实现方法。

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Little Pony and Permutation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 333 Accepted Submission(s): 215

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

Sample Output

  
   (1 2 5)(3 4)
(1)(2)(3)

Source

BestCoder Round #7

/*
就是置换水题，，，，
Time:2014-12-18 22:00
*/
#include<cstdio>
#include<cstring>
const int MAX=100000+10;
int main(){
    int n,i,j,k;
    int buf[MAX];
    while(scanf("%d",&n)!=EOF){
        for(i=1;i<=n;i++)
            scanf("%d",&buf[i]);
        for(i=1;i<=n;i++){
            while(buf[i]){
                printf("(%d",i);
                j=buf[i];buf[i]=0;
                while(buf[j]){
                    printf(" %d",j);
                    k=buf[j];
                    buf[j]=0;
                    j=k;
                }
                printf(")");
            }
        }puts("");
    }
return 0;
}