HDU 6184 Counting Stars

Problem Description

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

Input

There are no more than 300 test cases.
For each
test case, there are 2 positive integers n and m in the first line.

2≤n≤10⁵, 1≤m≤min(2×10⁵,n(n−1)²)
And then m lines follow, in each line there are two positive integers u and v, describing that this edge connects node u and node v.
1≤u,v≤n 
∑n≤3×10⁵,∑m≤6×10⁵

Output

For each test case, just output one integer--the number
of different "A-structure"s in one line.

Sample Input

4 5

1 2

2 3

3 4

4 1

1 3

4 6

1 2

2 3

3 4

4 1

1 3

2 4

Sample Output

1

6

题意：给定一张无向图，求有公共边的三元环对数。

Solution：

　　三元环裸题。

　　直接三元环计数，然后开一个桶记录一下每条边在多少个三元环中出现，最后的答案就是$\sum_\limits{i=1}^{i\leq m}{\frac{tot[i]*(tot[i]-1)}{2}}$。

代码：

/*Code by 520 -- 9.10*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define il inline
#define ll long long
#define RE register
#define For(i,a,b) for(RE int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(RE int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=100005,M=200005;
int n,m,to[M],net[M],h[N],cnt,tot[M],pre[N],vis[N],deg[N];
struct node{
    int u,v;
}e[M];
ll ans;

il void add(int u,int v){to[++cnt]=v,net[cnt]=h[u],h[u]=cnt;}

int main(){
    while(scanf("%d%d",&n,&m)==2){
        For(i,1,m) scanf("%d%d",&e[i].u,&e[i].v),deg[e[i].u]++,deg[e[i].v]++;
        For(i,1,m) {
            RE int u=e[i].u,v=e[i].v;
            if(deg[u]<deg[v]||deg[u]==deg[v]&&u>v) swap(u,v);
            add(u,v);
        }
        For(u,1,n){
            for(RE int i=h[u];i;i=net[i]) vis[to[i]]=u,pre[to[i]]=i;
            for(RE int i=h[u];i;i=net[i]){
                RE int v=to[i];
                for(RE int j=h[v];j;j=net[j]){
                    RE int w=to[j];
                    if(vis[w]==u) ++tot[i],++tot[j],++tot[pre[w]];
                }
            }
        }
        For(i,1,cnt) ans+=1ll*tot[i]*(tot[i]-1)/2;
        printf("%lld\n",ans);
        memset(h,0,sizeof(h)),memset(deg,0,sizeof(deg)),
        memset(tot,0,sizeof(tot)),memset(pre,0,sizeof(pre)),
        memset(vis,0,sizeof(vis)),cnt=0,ans=0;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/five20/p/9623592.html