判平行四边形的个数

// D题 判平行四边形的个数  忘记了数学方法
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
//const int MAXN =2e5+10;
const int N =332748118;

struct  edge
{
    int to;
    int next;
}edge[30009];
int head[30009];
int vis[3009];
int check[3009];

int main()
{
    int i,p,j,n,t;
    int m,cnt,a,b,x;
    scanf("%d%d",&n,&m);

    memset(head,-1,sizeof(head));
    cnt = 0;
    while(m--)
    {
        scanf("%d%d",&a,&b);
        edge[cnt].to = b;
        edge[cnt].next = head[a];
        head[a] = cnt++;
    }

    ans = 0;
    for(i = 1; i <= n; i++)
    {
        memset(vis,0,sizeof(vis));
        for(j = head[i]; j != -1; j = edge[j].next)
        {
            x = edge[j].to;
            for(p = head[x]; p != -1; p = edge[p].next)
            {
                if(edge[p].to != i)
                {
                    vis[edge[p].to] ++;
                }
            }
        }

        for(j = 1; j <= n; j++)
        {
            if(i != j)
            {
                ans += vis[j] * (vis[j] - 1) / 2;
            }
        }

    }


    printf("%d
", ans);
    return 0;
}