Codeforces--939A--Love Triangle

題目描述：
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 f i, where 1 ≤ f i ≤ n and f i ≠ 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.
輸入描述:
The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.

The second line contains n integers f 1, f 2, …, f n (1 ≤ f i ≤ n, f i ≠ i), meaning that the i-th plane likes the f i-th.
輸出描述:
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.
輸入：
5
2 4 5 1 3
5
5 5 5 5 1
輸出：
YES
NO
題意：
在幼兒園中有 n 個小朋友, 這些小朋友從 1 到 n 標號，每個小朋友都有一個特別想要一起玩的同伴，標號爲 i 的小朋友特別希望能夠和標號爲 fi 的小朋友一起玩, 其中 1 ≤ fi ≤ n 且 fi ≠ i.

如果存在三個小朋友 A, B, C 滿足 A 想和 B 玩, B 想和 C 玩，C 想和 A 玩，那麼這三個小朋友就被稱爲一個三人遊戲組，給出每個小朋友最想和哪個小朋友一起玩，你需要判斷其中是否存在一個三人遊戲組
題解：
直接暴力
代碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int maxn = 5000 + 5;
int a[maxn];

int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i = 1; i <= n; i ++) scanf("%d",&a[i]);
        bool flag = false;
        for(int i = 1; i <= n; i ++){
            if(i == a[a[a[i]]]){
                flag = true;
                break;
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}