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 integern(1<=n<=5000),the number of points. There aren 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，那么我们只要再找两个端点在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;}