poj1971Parallelogram Counting

链接

越来越感觉到了数学的重要性！。。

这题本来用以斜率和长度为key值进行hash不过感觉很麻烦还TLE了。。

最后知道中点一样的话就可以组成平行四边形，初中数学就可以了。。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define N 1010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct point
{
    int x,y;
} p[N],o[N*N];

bool cmp(point a,point b)
{
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}
int main()
{
    int t,n,i,j;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int g = 0;
        for(i = 1 ; i<= n ; i++)
        {
            for(j = i+1 ; j <= n ;j++)
            {
                o[++g].x = (p[i].x+p[j].x);
                o[g].y = (p[i].y+p[j].y);
               // cout<<o[g].y<<endl;
            }
        }
        LL ans = 0;
        sort(o+1,o+g+1,cmp);
        int e = 1;
        for(i = 2; i <= g ;i++)
        {
            if(o[i].x-o[i-1].x==0&&o[i].y-o[i-1].y==0)
            {
                e++;
               // cout<<o[i].x<<" ,"<<o[i-1].x<<" "<<o[i].y<<" "<<o[i-1].y<<endl;
            }
            else
            {
                ans+=(LL)(e-1)*e/2;
                e = 1;
            }
        }
        ans+=(LL)(e-1)*e/2;
        cout<<ans<<endl;
    }
    return 0;
}