UVA 10574

Given n points on the XY plane, count how many regular rectangles are formed. A rectangle is regular if and only if its sides are all parallel to the axis.
Input
The first line contains the number of tests t (1 ≤ t ≤ 10). Each case contains a single line with a
positive integer n (1 ≤ n ≤ 5000), the number of points. There are n lines follow, each line contains 2
integers x, y (≤ x, y ≤ 109
) indicating the coordinates of a point.
Output
For each test case, print the case number and a single integer, the number of regular rectangles found.
Sample Input
2
5
0 0
2 0
0 2
2 2
1 1
3
0 0
0 30
0 900
Sample Output
Case 1: 1
Case 2: 0

题意：给你n个点 ，问你这些点能够组成多少个 长宽和坐标轴平行的 矩形

题解：按照x排序，在y轴平行下，选择不同直线组合，满足两y轴上的点相等就是一种，  对于一堆相等的 我们用组合数就好了

//meek
///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair

const int N=5005;
const ll INF = 1ll<<61;
const int inf = 1<<31;
const int mod= 1000000007;
const int M = 1000000;


struct ss{
  int x,y;
}a[N],p[N*N];
int cnt;
int cmp(ss s1,ss s2) {
    if(s1.x == s2.x) return s1.y<s2.y;
    return s1.x<s2.x;
}
void init() {
  cnt = 0;mem(p);
}
ll solve() {
    ll ans = 0;
  for(int i = 0;i < cnt; ) {
    int now = i+1;
    while(p[i].x == p[now].x && p[i].y == p[now].y) now++;
    ll c = now - i;
    if(c>=2) ans += c*(c-1)/2;
    i = now;
  }
  return ans;
}
int main() {
    int T,n,x,y,cas = 1;
    scanf("%d",&T);
    while(T--) {
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++) {
            scanf("%d%d",&x,&y);
            a[i].x = x;
            a[i].y = y;
        }
        sort(a+1,a+n+1,cmp);
        cnt = 0;
        for(int i=1;i<=n;i++) {
            for(int j=i+1;j<=n;j++) {
                if(a[i].x != a[j].x) {
                   break; 
                }
                p[cnt].x = a[i].y;
                p[cnt].y = a[j].y;
                cnt++;
            }
        }
        printf("Case %d: ",cas++);
        sort(p,p+cnt,cmp);
        printf("%lld
",solve());
    }
    return 0;
}