题目链接：点击打开链接

题意：

给定n个点m条边的无向图

问图中是否存在 有且仅有一个简单环和一些树，且这些树的root都在这个简单环上。

瞎写了个点双。。==

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
#include <set>
using namespace std;
#define N 105
#define M 100005
#define inf 10000000
struct Edge{
	int from,to,next;
}edge[2*M];
int head[N],edgenum;
int Low[N],DFN[N],Stack[N];//Belong数组的值是1~block
int Index,top;
int Belong[N],block;//新图的连通块标号（1~block）
bool Instack[N];
int bridge; //割桥数量
vector<int> bcc[N];
void addedge(int u,int v){
	Edge E={u,v,head[u]}; edge[edgenum]=E; head[u] = edgenum++;
	Edge E2={v,u,head[v]};edge[edgenum]=E2;head[v] = edgenum++;
}
void Tarjan(int u,int pre){
	int v;
	Low[u] = DFN[u] = ++Index;
	Stack[top++] = u;
	Instack[u] = true;
	for(int i = head[u]; ~i ;i = edge[i].next){
		v = edge[i].to;
		// 如果重边有效的话下面这句改成: if(v == pre && pre_num == 0){pre_num++;continue;} pre_num在for上面定义 int pre_num=0;
		if( v == pre )continue;
		if( !DFN[v] ){
			Tarjan(v,u);
			if(Low[u] > Low[v])Low[u] = Low[v];
			if(Low[v] > Low[u])
				bridge++;
		}
		else if(Instack[v] && Low[u] > DFN[v])Low[u] = DFN[v];
	}
	if(Low[u] == DFN[u]){
		block++;
		bcc[block].clear();
		do{
			v = Stack[--top];
			Instack[v] = false;
			Belong[v] = block;
			bcc[block].push_back(v);
		}while( v != u );
	}
}
void work(int l, int r){
	memset(DFN,0,sizeof(DFN));
	memset(Instack,false,sizeof(Instack));
	Index = top = block = bridge = 0;
	for(int i = l; i <= r; i++)if(!DFN[i])Tarjan(i,i);
}
vector<int>G[N];
void suodian(){
	for(int i = 1; i <= block; i++)G[i].clear();
	for(int i = 0; i < edgenum; i+=2){
		int u = Belong[edge[i].from], v = Belong[edge[i].to];
		if(u==v)continue;
		G[u].push_back(v), G[v].push_back(u);
	}
}
void init(){edgenum = 0; memset(head,-1,sizeof(head));}

int n, m, vis[N], siz[N], f[N];
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
void Union(int x, int y){
    int fx = find(x), fy = find(y);
    if(fx==fy)return;
    if(fx>fy)swap(fx,fy); f[fx] = fy;
}
bool dfs(int u){
	vis[u] = 1;
	bool ok = true;
	for(int i = 0; i < G[u].size(); i++)
	{
		int v = G[u][i]; if(vis[v])continue;
		if(bcc[v].size()>1)ok = false;
		if(dfs(v)==false)ok = false;
	}
	return ok;
}
set<int>myset;
bool solve(){
	int i, u, v;
	init();
	for(i = 1; i <= n; i++)f[i] = i;
	while(m--){
		scanf("%d %d",&u,&v);
		addedge(u, v);
		Union(u,v);
	}
	myset.clear();
	for(i = 1; i<= n; i++)myset.insert(find(i));
	if(myset.size()>1)return false;
	work(1, n);
	suodian();
	memset(vis, 0, sizeof vis);
	memset(siz, 0, sizeof siz);
	for(i = 0; i < edgenum; i+=2)
		siz[u]+= (Belong[edge[i].from]==Belong[edge[i].to]);

	for(i = 1; i <= block; i++) if(!vis[i] && bcc[i].size()>1 && bcc[i].size()==siz[i] && G[i].size()>=3)
		if(dfs(i))return true;

	return false;
}
int main(){
	while(~scanf("%d %d", &n, &m))
		solve() ? puts("FHTAGN!") : puts("NO");
	return 0;
}
/*
6 6
1 2
2 3
3 1
4 5
5 6
6 4

*/

Codeforces 104C Cthulhu dfs暴力 || 双连通缩点