Codeforces 103B. Cthulhu 寻找奈亚子

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本文深入探讨了一种独特且复杂的图论问题，涉及到Cthulhu这一神秘生物与全球超级大国Pentagon之间的隐秘联系。通过分析一个含有n个顶点和m条边的无向图，揭示了如何判断该图是否能被视作Cthulhu的形象。Cthulhu形象的解析不仅涉及到图的结构分析，还融入了抽象的数学逻辑和丰富的想象力。文章详细介绍了判断条件和算法过程，旨在帮助读者理解这一看似荒诞却又充满逻辑的数学问题，同时揭示了隐藏在科学与幻想交织背后的深层思考。

...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...

Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with n vertices and m edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.

To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.

It is guaranteed that the graph contains no multiple edges and self-loops.

$\text{[math]}$

Input

The first line contains two integers — the number of vertices n and the number of edges m of the graph (1 ≤ n ≤ 100, 0 ≤ m ≤ $\text{[math]}$ ).

Each of the following m lines contains a pair of integers x and y, that show that an edge exists between vertices x and y (1 ≤ x, y ≤ n, x ≠ y). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.

Output

Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.

Sample test(s)

input
6 6
6 3
6 4
5 1
2 5
1 4
5 4

output
FHTAGN!

input
6 5
5 6
4 6
3 1
5 1
1 2

output
NO

Note

Let us denote as a simple cycle a set of v vertices that can be numbered so that the edges will only exist between vertices number 1 and2, 2 and 3, ..., v - 1 and v, v and 1.

A tree is a connected undirected graph consisting of n vertices and n - 1 edges (n > 0).

A rooted tree is a tree where one vertex is selected to be the root.

奈亚子只有一个环，环上每个点都是一棵树。。。

#include <iostream>
#include <cstring>

using namespace std;

int n,m;
int a[111][111];
int p[111]= {0};
int dis[111]= {0};
bool v[111]= {0};
int cnt=0;

bool find_cycle(int i,int papa)
{
    //cerr<<i<<" ------- "<<papa<<endl;
    v[i]=true;
    for (int j=1; j<=n; j++)
    {
        if (j!=papa&&i!=j&&a[i][j]>0)
        {
            if (!v[j])
            {
                p[cnt++]=j;
                if (find_cycle(j,i)) return true;
                cnt--;
            }
            else
            {
                p[cnt++]=j;
                return true;
            }
        }
    }
    return false;
}

bool dfs(int i,int papa)
{
    v[i]=true;
    for (int j=1;j<=n;j++)
    {
        if (i!=j&&j!=papa&&a[i][j])
        {
            if (papa==0&&dis[j]==0) continue;
            if (v[j]||dis[j]==0)
            {
                return true;
            }
            else
            {
                if (dfs(j,i)) return true;
            }
        }
    }
    return false;
}

void len_dfs(int i)
{
    v[i]=true;
    for (int j=1;j<=n;j++)
    {
        if (!v[j]&&a[i][j]>0) len_dfs(j);
    }
}

int main()
{
    memset(dis,-1,sizeof(dis));
    cin>>n>>m;
    for (int i=1; i<=m; i++)
    {
        int x,y;
        cin>>x>>y;
        a[x][y]=a[y][x]=1;
    }
    p[cnt++]=1;
    memset(v,0,sizeof(v));
    bool ret=find_cycle(1,0);
    if (ret)
    {
        dis[p[cnt-1]]=0;
        for (int i=cnt-2; i>=0; i--)
        {
            if (p[i]==p[cnt-1]) break;
            dis[p[i]]=0;
        }
        //ready_dfs
        for (int i=1;i<=n;i++)
        {
            if (dis[i]==0)
            {
                memset(v,0,sizeof(v));
                if (dfs(i,0))
                {
                    ret=false;
                    break;
                }
            }
        }
        //dfs_over
    }

    memset(v,0,sizeof(v));
    len_dfs(1);
    for (int i=1;i<=n;i++)
    {
        if (!v[i]) ret=false;
    }
    if (ret) cout<<"FHTAGN!"<<endl;
    else cout<<"NO"<<endl;

    return 0;
}