题目链接：http://codeforces.com/problemset/problem/939/A
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.
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 «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.
5
2 4 5 1 3
YES
5
5 5 5 5 1
NO
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles.
思路：题目大意就是1号喜欢2号，2号喜欢3号，3号喜欢1号，如何去表示呢？用数组来下标来表示，例如a[1]=2，表示1号喜欢2号，同理a[2] = 3,表示2号喜欢3号，a[3] = 1,表示3号喜欢1号。现在的遇到的困难是，如何去表示这三者的关系。请先看AC代码：
#include<iostream> using namespace std; int n,a[5001]; int main() { while(cin >> n) { int flag = 0;//设一个标记 for(int i = 1;i <= n;i++) cin >> a[i]; for(int i = 1;i <= n;i++) if(a[a[a[i]]] == i)//只要有满足条件的马上跳出循环，立刻结束 flag = 1; if(flag == 1) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }
a[1]=2,a[2]=3,a[3]=1，这是一组满足条件的三角恋关系，我们拿这个例子来分析。a[1]=2说明1号喜欢2号，我们马上判断2号喜欢的是谁,我们想要知道2号喜欢谁，把a[1]=2（1号喜欢2号）中的2号放入数组a中，即a[a[1]]，因为a[1]=2，a[a[1]]等价于a[2]，这表示的是2号喜欢的是谁，同理，a[2]=3，如何表示3号喜欢谁呢?再把a[a[1]]放入数组a中，即a[a[a[1]]]（即为3号喜欢的是谁），判断3号是否喜欢1号，如果是，则三者满足三角恋的条件，否则不满足，继续判断。好好理解下，理解后就不难了。
原文：https://www.cnblogs.com/biaobiao88/p/11846203.html