HDU6184 Counting Stars

题意：

Little A is an astronomy lover, and he has found that the sky was so beautiful!
So he is counting stars now!
There are n stars in the sky, and little A has connected them by m non-directional edges.
It is guranteed that no edges connect one star with itself, and every two edges connect different pairs of stars.
Now little A wants to know that how many different "A-Structure"s are there in the sky, can you help him?
An “A-structure” can be seen as a non-directional subgraph G, with a set of four nodes V and a set of five edges E.
If V=(A,B,C,D) and E=(AB,BC,CD,DA,AC), we call G as an “A-structure”.
It is defined that “A-structure” G1=V1+E1 and G2=V2+E2 are same only in the condition that V1=V2 and E1=E2.

数据范围：

$n\le 10^5,m\le 2\ast 10^5$

Analysis：

翻译不是这题的重点，我们来讲讲这题真正要求的东西，其实就是懒得翻译，英文不好。
事实上就是让我们三元环计数，找出一条边存在于多少个三元环中。
我们考虑一种暴力：枚举一个点，枚举所有出边的点打上标记，再枚举出边，然后枚举出边的点的出边，看出边的点是否有标记，语文功底太差，超级绕 。我们设一个点的出度为$ou{t}_v$，不难发现这样的复杂度为${\sum }_{e_{u,v}}ou{t}_u+ou{t}_v$。让每条边只会被枚举两次。
考虑平衡出度，我们强制让度数小的连向度数大的点(有向边)，度数相同按编号有序连边。
不难发现会是一张$Dag$，因为如果有环的话，编号的偏序关系会矛盾。原先的三元环仍然可以如上枚举，且仅会被枚举一次，因为必定会有且仅有一个点入度为$2$。
这样复杂度变为${\sum }_{e_{u,v}}ou{t}_v$，那么每个点出度为多少呢，是不会超过$\sqrt{m}$的。
考虑反证：若一个点出度超过$\sqrt{m}$，因为连出去的点出度都大于其，所以至少$\sqrt{m}$个点，每个点至少$\sqrt{m}$条出边，出边总数大于$m$ ，所以与条件矛盾。
那么总复杂度为$O(m\sqrt{m})$。比较优秀，常数也比较小。
至于统计答案，只要算出每条边存在于多少个三元环，然后$C_{cnt}^2$加起来就好。

Code：

# include
# include
# include
using namespace std;
const int N = 1e5 + 5;
typedef long long ll;
int st[N],to[N << 1],nx[N << 1],id[N << 1];
int vis[N][2],D[N],e[N << 1][2],cnt[N << 1];
int n,m,tot;
inline void add(int u,int v,int w)
{ to[++tot] = v,nx[tot] = st[u],id[tot] = w,st[u] = tot; }
int main()
{
	while (~scanf("%d%d",&n,&m))
	{
		memset(vis,0,sizeof(vis)),memset(st,0,sizeof(st)),memset(cnt,0,sizeof(cnt)),memset(D,0,sizeof(D)),tot = 0;
		for (int i = 1 ; i <= m ; ++i) scanf("%d%d",&e[i][0],&e[i][1]),++D[e[i][0]],++D[e[i][1]];
		for (int i = 1 ; i <= m ; ++i)
		{
			if (e[i][0] > e[i][1]) swap(e[i][0],e[i][1]);
			if (D[e[i][0]] > D[e[i][1]]) swap(e[i][0],e[i][1]);
			add(e[i][0],e[i][1],i);
		}
		for (int i = 1 ; i <= n ; ++i)
		{ 
			for (int j = st[i] ; j ; j = nx[j]) vis[to[j]][0] = i,vis[to[j]][1] = j;
			for (int j = st[i] ; j ; j = nx[j])
				for (int k = st[to[j]] ; k ; k = nx[k])
				if (vis[to[k]][0] == i) ++cnt[id[j]],++cnt[id[k]],++cnt[vis[to[k]][1]];
		}
		ll ans = 0;
		for (int i = 1 ; i <= m ; ++i) if (cnt[i] > 1) ans += (ll)(cnt[i] - 1) * cnt[i] / 2;
		printf("%lld\n",ans);
	}
	return 0;
}