Answer to Question #209842 in C++ for Jhonny harris

Question

Form Rectangle from Triangles A rectangle shall be formed from many triangles. Given vertices of 'n' triangles ,with some triangles among them that can form a rectangle, design an algorithm and write a C++ code to determine the triangles that can form a rectangle. Print the vertices of the triangle in sorted order. That is triangles should also be sorted and the vertices in a triangle also should be sorted. A rectangle may contain four triangles as shown in figure below: For example, consider six triangles with the following vertices: Triangle 1 – (10, 6), (10, 16), (5, 6) Triangle 2 - (25, 8), (20, 4), (28, 6) Triangle 3 – (0, 6), (0, 16), (5, 6) Triangle 4 – (0, 16), (10, 16), (5, 6) Triangle 5 – (15,10), (20, 15), (25, 8) Triangle 6 – (0, 6 ), (5, 6 ), (10, 6) Triangle 1,3 and 4 can form a rectangle.

Answer 1

#include <bits/stdc++.h> 
using namespace std; 
#define NUMBER_OF_TRIANGlES 4 
  


struct Triangles_Coordinates
{ 
    int x1, y1; 
    int x2, y2; 
   
}; 
  


int getDistance(pair<int, int> x, pair<int, int> y) 
{ 
    return (x.first - y.first)*(x.first - y.first) + 
        (x.second - y.second)*(x.second - y.second); 
} 
  


bool isRectangle(Triangles_Coordinates triangle[]) 
{ 
    set< pair<int, int> > code; 
  
    
    for (int n = 0; n < NUMBER_OF_TRIANGlES; n++) 
    { 
        code.insert(make_pair(triangle[n].x1, triangle[n].y1)); 
        code.insert(make_pair(triangle[n].x2, triangle[n].y2)); 
       
    } 
  
   
    if (code.size() != 4) 
        return false; 
  
  
    set<int> distance_array; 
  
    
    for (auto x=code.begin(); x!=code.end(); x++) 
        for (auto y=code.begin();y!=code.end(); y++) 
            if (*x != *y) 
                distance_array.insert(getDistance(*x, *y)); 
   
    if (distance_array.size() > 3) 
        return false; 
  
     
    int distances[3]; 
    int r = 0; 
    for (auto code1 = distance_array.begin(); code1 != distance_array.end(); code1++) 
        distances[r++] = *code1; 
  
   
    if (distance_array.size() == 2) 
    return (2*distances[0] == distances[1]); 
  
    
    return (distances[0] + distances[1] == distances[2]); 
} 
  


int main() 
{ 
    Triangles_Coordinates triangle[] = 
    { 
        {4, 2, 7, 5}, 
        {2, 4, 4, 2}, 
        {2, 4, 5, 7}, 
        {5, 7, 7, 5} 
    }; 
  
    (isRectangle(triangle))?cout << "Yes\n":cout << "No\n"; 
}