ZOJ - 2795 Ambiguous permutations-优快云博客

本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/47081447

本文深入探讨了模糊排列的概念，通过实例展示了如何识别一个排列是否为模糊排列，并提供了一个简洁有效的检测算法。

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Ambiguous permutations

Time Limit: 10000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

Description

Some programming contest problems are really tricky: not only do they require a different output format from what you might have expected, but also the sample output does not show the difference. For an example, let us look at permutations.
A permutation of the integers 1 to n is an ordering of these integers. So the natural way to represent a permutation is to list the integers in this order. With n = 5, a permutation might look like 2, 3, 4, 5, 1.
However, there is another possibility of representing a permutation: You create a list of numbers where the i-th number is the position of the integer i in the permutation. Let us call this second possibility an inverse permutation. The inverse permutation for the sequence above is 5, 1, 2, 3, 4.
An ambiguous permutation is a permutation which cannot be distinguished from its inverse permutation. The permutation 1, 4, 3, 2 for example is ambiguous, because its inverse permutation is the same. To get rid of such annoying sample test cases, you have to write a program which detects if a given permutation is ambiguous or not.

Input Specification

The input contains several test cases.
The first line of each test case contains an integer n (1 ≤ n ≤ 100000). Then a permutation of the integers 1 to n follows in the next line. There is exactly one space character between consecutive integers. You can assume that every integer between 1 and n appears exactly once in the permutation.
The last test case is followed by a zero.

Output Specification

For each test case output whether the permutation is ambiguous or not. Adhere to the format shown in the sample output.

Sample Input

Sample Output

ambiguous
not ambiguous
ambiguous

Source

University of Ulm Local Contest 2005

分析：

水题。

ac代码：

#include <iostream>
#include<cstdio>
using namespace std;
const int maxn=100005;
int a[maxn];
int main()
{
    int n,i;
    bool flag;
    while(scanf("%d",&n)&&n)
    {
        flag=false;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=1;i<=n;i++)
        {
            if(i!=a[a[i]])
            {
                flag=true;
            }
        }
        if(flag)
        printf("not ambiguous\n");
        else
        printf("ambiguous\n");
    }
    return 0;
}