Counting Stars HDU - 6184（无向图三元环计数）

Counting Stars HDU - 6184
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

无向图三元环详解
这里需要一些小小的改动
题目中要求的A结构其实是两个具有一条公共边的三元环
求这样的个数
根据算法我们一开始就是先枚举一条边，这样我们就以第一条边为公共边，找出三元环的个数，然后$C_{sum}^{2}$从这n个中选2个即可，也就是$\frac{sum\cdot(sum-1)}{2}$
code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
vector<int>G[maxn];
set<ll>st;
int vis[maxn],link[maxn],out[maxn];
int main(){
    ll ans,sum;
    int n,m,i,j,k,x,y,z,B;
    while(scanf("%d%d",&n,&m) != EOF){
        //初始化
        B = sqrt(m);
        st.clear();
        for(i = 1; i <= n; i++){
            G[i].clear();
            vis[i] = out[i] = link[i] = 0;
        }
        for(i = 1; i <= m; i++){
            scanf("%d%d",&x,&y);
            G[x].push_back(y);
            out[x]++;
            G[y].push_back(x);
            out[y]++;
            st.insert((ll)x*n+y);
            st.insert((ll)y*n+x);
        }
        ans = 0;
        for(i = 1; i <= n; i++){
            x = i;
            vis[x] = 1;
            for(j = 0; j < G[x].size(); j++)
                link[G[x][j]] = x;
            for(j = 0; j < G[x].size(); j++){
                sum = 0;
                y = G[x][j];
                if(vis[y])
                    continue;
                if(out[y] <= B){
                    for(k = 0; k < G[y].size(); k++){
                        z = G[y][k];
                        if(link[z] == x)
                            sum++;
                    }
                }
                else{
                    for(k = 0; k < G[x].size(); k++){
                        z = G[x][k];
                        if(st.find((ll)z*n+y) != st.end())
                            sum++;
                    }
                }
                ans += sum * (sum - 1) / 2;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}