Codeforces Round #464 (Div. 2) A. Love Triangle

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本文介绍了一个简单的算法，用于判断一组给定的关系中是否存在三角恋。通过三次查询喜欢的对象来判断是否存在闭环，进而确定是否存在三角恋。

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As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ i.

We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.

The second line contains n integers f1, f2, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-th.

Output

Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».

You can output any letter in lower case or in upper case.

Examples
inputCopy
5
2 4 5 1 3
output
YES
inputCopy
5
5 5 5 5 1
output
NO

题意：给你n个数，第i个数代表：第i个人喜欢第a[i]个人，问你是否存在三角恋。

思路：第i个下标对应的值a[i]即为喜欢的人，所以很容易发现如果a[a[a[i]]]==i，即为三角恋。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[5050],n,flag;
int main()
{
    while(~scanf("%lld",&n))
    {
        for(ll i=1;i<=n;i++)scanf("%lld",&a[i]);
        for(ll i=1;i<=n;i++)
        {
            if(a[a[a[i]]]==i)flag=1;   //三角恋判定
        }
        if(flag==1)printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}