查找可以从给定的一组坐标形成的矩形数量
给定一个数组arr[][]由一对表示坐标的整数组成。任务是计算使用给定坐标可以形成的矩形的总数。
例子：
Input: arr[][] = {{0, 0}, {0, 1}, {1, 0}, {1, 1}, {2, 0}, {2, 1}, {11, 11}}
Output: 3
Explanation: Following are the rectangles formed using given coordinates.
First Rectangle (0, 0), (0, 1), (1, 0), (1, 1)
Second Rectangle (0, 0), (0, 1), (2, 0), (2, 1)
Third Rectangle (1, 0), (1, 1), (2, 0), (2, 1)
Input: arr[][] = {{10, 0}, {0, 10}, {11, 11}, {0, 11}, {12, 10}}
Output: 0
Explanation: No Rectangles can be formed using these co-ordinates
方法：这个问题可以通过使用矩形和哈希映射的属性来解决。如果一个矩形的两个坐标是已知的，那么剩下的两个坐标就可以很容易地确定。对于每一对坐标，找到其他两个可以形成矩形的坐标。
下面是上述方法的实现。
C++
// C++ program for above approach
#include
using namespace std;
// Function to find number of possible rectangles
int countRectangles(vector >& ob)
{
// Creating TreeSet containing elements
set > it;
// Inserting the pairs in the set
for (int i = 0; i < ob.size(); ++i) {
it.insert(ob[i]);
}
int ans = 0;
for (int i = 0; i < ob.size(); ++i)
{
for (int j = 0; j < ob.size(); ++j)
{
if (ob[i].first != ob[j].first
&& ob[i].second != ob[j].second)
{
// Searching the pairs in the set
if (it.count({ ob[i].first, ob[j].second })
&& it.count(
{ ob[j].first, ob[i].second }))
{
// Increase the answer
++ans;
}
}
}
}
// Return the final answer
return ans / 4;
}
// Driver Code
int main()
{
int N = 7;
vector > ob(N);
// Inserting the pairs
ob[0] = { 0, 0 };
ob[1] = { 1, 0 };
ob[2] = { 1, 1 };
ob[3] = { 0, 1 };
ob[4] = { 2, 0 };
ob[5] = { 2, 1 };
ob[6] = { 11, 23 };
cout << countRectangles(ob);
return 0;
}
// This code is contributed by rakeshsahni Java
// Java program for above approach
import java.io.*;
import java.util.*;
class Main {
// Creataing pair class and
// implements comparable interface
static class Pair implements Comparable {
int first;
int second;
Pair(int first, int second)
{
this.first = first;
this.second = second;
}
// Changing sorting order of the pair class
@Override
public int compareTo(Pair o)
{
// Checking the x axis
if (this.first == o.first) {
return this.second - o.second;
}
return this.first - o.first;
}
}
// Function to find number of possible rectangles
static int countRectangles(Pair ob[])
{
// Creating TreeSet containing elements
TreeSet it = new TreeSet<>();
// Inserting the pairs in the set
for (int i = 0; i < ob.length; ++i) {
it.add(ob[i]);
}
int ans = 0;
for (int i = 0; i < ob.length; ++i) {
for (int j = 0; j < ob.length; ++j) {
if (ob[i].first != ob[j].first
&& ob[i].second != ob[j].second) {
// Searching the pairs in the set
if (it.contains(new Pair(ob[i].first,
ob[j].second))
&& it.contains(new Pair(
ob[j].first, ob[i].second))) {
// Increase the answer
++ans;
}
}
}
}
// Return the final answer
return ans / 4;
}
// Driver Code
public static void main(String[] args)
{
int N = 7;
Pair ob[] = new Pair[N];
// Inserting the pairs
ob[0] = new Pair(0, 0);
ob[1] = new Pair(1, 0);
ob[2] = new Pair(1, 1);
ob[3] = new Pair(0, 1);
ob[4] = new Pair(2, 0);
ob[5] = new Pair(2, 1);
ob[6] = new Pair(11, 23);
System.out.print(countRectangles(ob));
}
} Python3
# Python program for above approach
# Function to find number of possible rectangles
def countRectangles(ob):
# Creating TreeSet containing elements
it = set()
# Inserting the pairs in the set
for i in range(len(ob)):
it.add(f"{ob[i]}");
ans = 0;
for i in range(len(ob)):
for j in range(len(ob)):
if (ob[i][0] != ob[j][0] and ob[i][1] != ob[j][1]):
# Searching the pairs in the set
if (f"{[ob[i][0], ob[j][1]]}" in it and f"{[ob[j][0], ob[i][1]]}" in it):
# Increase the answer
ans += 1
# Return the final answer
return int(ans / 4);
# Driver Code
N = 7;
ob = [0] * N
# Inserting the pairs
ob[0] = [0, 0];
ob[1] = [1, 0];
ob[2] = [1, 1];
ob[3] = [0, 1];
ob[4] = [2, 0];
ob[5] = [2, 1];
ob[6] = [11, 23];
print(countRectangles(ob));
# This code is contributed by Saurabh JaiswalJavascript
输出：
3时间复杂度：O(N 2 )，其中 N 是数组的大小。
辅助空间：O(N)，其中 N 是数组的大小。