UVA - 10574 Counting Rectangles

Description

Problem H

Counting Rectangles

Input: Standard Input

Output:Standard Output

Time Limit: 3Seconds

 

Given n points on the XY plane, count how many regular rectanglesare formed. A rectangle is regular if and only if its sides are all parallel tothe axis.

 

Input

Thefirst 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 2integers x, y (0<=x, y<=10⁹)indicating the coordinates of a point.

 

Output

Foreach test case, print the case number and a single integer, the number ofregular rectangles found.

 

SampleInput                            Output for Sample Input

2 5 0 0 2 0 0 2 2 2 1 1 3 0 0 0 30 0 900

Case 1: 1 Case 2: 0

题意：给定平面上的n个点，统计它们能组成多少个边平行于坐标轴的矩形

思路：题目要求平行于坐标轴，那么我们先找同一x轴的两个点组成的线段，保存两个点的y轴坐标，那么我们仅仅要再找两个端点在y轴平行的这样就能找到矩形了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 5010;

struct Point {
	int x, y;
	bool operator< (const Point &a) const {
		if (x != a.x)
			return x < a.x;
		return y < a.y;
	}
} p[maxn];

struct Edge {
	int y1, y2;
	Edge() {}
	Edge(int y1, int y2) {
		this->y1 = y1;
		this->y2 = y2;
	}
	bool operator <(const Edge &a) const {
		if (y1 != a.y1)
			return y1 < a.y1;
		return y2 < a.y2;
	}
} e[maxn*maxn];
int n;

int main() {
	int t;
	int cas = 1;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++) 
			scanf("%d%d", &p[i].x, &p[i].y);
		sort(p, p+n);
		int num = 0;
		for (int i = 0; i < n; i++) 
			for (int j = i+1; j < n; j++) {
				if (p[i].x != p[j].x)
					break;
				e[num++] = Edge(p[i].y, p[j].y);
			}
		sort(e, e+num);
		int tmp = 1, ans = 0;
		for (int i = 1; i < num; i++) {
			if (e[i].y1 == e[i-1].y1 && e[i].y2 == e[i-1].y2)
				tmp++;
			else {
				ans += tmp * (tmp-1) / 2;
				tmp = 1;
			}
		}
		ans += tmp * (tmp-1) / 2;
		printf("Case %d: %d\n", cas++, ans);
	}
	return 0;
}