Codeforces 939A Love Triangle（三角恋问题）

最新推荐文章于 2021-11-15 22:27:18 发布

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本文介绍了一道关于是否存在三角恋情况的问题，并提供了一个简洁的C++解决方案。通过遍历每一对飞机的好感关系来判断是否存在三角恋，最终输出判断结果。

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题目：

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
Note
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.

题意：

先输入一个n，然后数组有n个数，表示喜欢谁，如a[1]=2；意思就是1号喜欢的是2号。问你存不存在三角恋的情况存在输出YES 否则输出NO

思路：

判断第一个人喜欢的那个人喜欢的人喜欢的是不是自己。

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>

using namespace std;
const int N=1e5+10;
int a[N];

int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            int x,y;
            x=a[i];//自己喜欢的人
            y=a[x];//自己喜欢的人喜欢的人
           if(a[y]==i&&x!=i&&x!=y&&i!=y)
           //自己喜欢的人不能是自己  自己喜欢的人不能喜欢自己 自己喜欢的人的喜欢的人不能是自己
            flag=1;
            if(flag==1)
                break;
        }
        if(flag)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;

}