【dfs找環】Codeforces Beta Round #80 (Div. 1 Only)——B. Cthulhu

題意是否存在多棵樹，它們的根形成一個環，簡單來說就是求無向圖中是否存在只有一個環，而且各點連通。

方法：樹中只有一環的性質是邊數=點數，滿足這個還不能下定論，因為有可能存在多個環野滿足這樣的條件，所以接下來就dfs判連通性。

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <time.h>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define LL long long#define MIN INT_MIN#define MAX INT_MAX#define PI acos(-1.0)#define FRE freopen ("input.txt","r",stdin)#define FF  freopen ("output.txt","w",stdout)#define eps 1e-8#define N 105int g[N][N];bool vis[N];int cnt;int n;void dfs(int x) {    int i,j;    vis[x] = 1;    cnt++;    for (i = 1; i <= n; i++) {        if (!vis[i] && g[x][i]) {            dfs(i);        }    }}int main(){    int m;    while(scanf("%d%d",&n,&m) != -1) {        int i,j,k;        int mm = m;        memset(vis,0,sizeof(vis));        while (m--) {            int a,b;            scanf("%d%d",&a,&b);            g[a][b] = g[b][a] = 1;        }        if (n == mm) {            cnt = 0;            dfs(1);            if(cnt == n) puts("FHTAGN!");            else puts("NO");        }        else puts("NO");    }    return 0;}