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Area of the nearest boxes

Tags:

c++

algorithm

This question is similar to my previously asked this question . But instead of returning only the number of nearest boxes, I would like to find the area of corresponding boxes.

Details: Suppose I have a set of coordinates of boxes like this-

#Rect    x1      y1          x2       y2         area

1     0.0000   0.0000      0.8147   0.1355      0.1104
2     0.8147   0.0000      1.0000   0.1355      0.0251
3     0.8147   0.1355      0.9058   0.8350      0.0637
4     0.0000   0.1355      0.1270   0.9689      0.1058
5     0.9058   0.1355      0.9134   0.2210      0.0006
6     0.9058   0.8350      1.0000   1.0000      0.0155
7     0.8147   0.8350      0.9058   1.0000      0.0150
8     0.1270   0.1355      0.6324   0.3082      0.0873
9     0.1270   0.9689      0.8147   1.0000      0.0214
10    0.0000   0.9689      0.1270   1.0000      0.0040
11    0.9134   0.1355      1.0000   0.2210      0.0074
12    0.9134   0.2210      1.0000   0.8350      0.0532
13    0.9058   0.2210      0.9134   0.8350      0.0047
14    0.6324   0.1355      0.8147   0.3082      0.0315
15    0.6324   0.3082      0.8147   0.9689      0.1205
16    0.1270   0.3082      0.6324   0.9689      0.3339

Suppose these coordinates split an unit square into sub-rectangles like this picture- enter image description here

Now this code returns the value of the number of nearest boxes, but it fails to returns the area of that boxes. Here is my code-

#include <iostream>
#include <cstdlib>
#include <vector>
#include <stdio.h>

using namespace std;

class Rect {
  public:
  double x1, x2, y1, y2, area; 

  Rect(double X1, double Y1, double X2, double Y2, double Area) {
    if (X1 < X2) {
      x1 = X1; x2 = X2;
    } else {
      x2 = X1; x1 = X2;
    }
    if (Y1 < Y2) {
      y1 = Y1; y2 = Y2;
    } else {
      y2 = Y1; y1 = Y2;
    } Area =area;

  }

  bool NearestBox(Rect rect) {

    //for x-axis
    if (x1 == rect.x2 || x2 == rect.x1) {     

      if (y1 >= rect.y1 && y1 < rect.y2) {
        return true;
      }
      if (y2 > rect.y1 && y2 <= rect.y2) {
        return true;
      }
    }              

    // for y-axis    

    if (y1 == rect.y2 || y2 == rect.y1) {
      if (x1 >= rect.x1 && x1 < rect.x2) {
        return true;
      }
      if (x2 > rect.x1 && x2 <= rect.x2) {
        return true;
      }
    }

    return false;  

  }
};

int main() {

  vector<Rect> rects;     
                //Rect(  x1 ,  y1  ,   x2  ,  y2   ,  area) 
  rects.push_back(Rect(0.0000,0.0000, 0.8147,0.1355, 0.1104));
  rects.push_back(Rect(0.8147,0.0000, 1.0000,0.1355, 0.0251));

  rects.push_back(Rect(0.8147,0.1355, 0.9058,0.8350, 0.0637));
  rects.push_back(Rect(0.0000,0.1355, 0.1270,0.9689, 0.1058 ));

  rects.push_back(Rect(0.9058,0.1355, 0.9134,0.2210, 0.0006));
  rects.push_back(Rect(0.9058,0.8350, 1.0000,1.0000, 0.0155));
  rects.push_back(Rect(0.8147,0.8350, 0.9058,1.0000, 0.0150));



  rects.push_back(Rect(0.1270,0.1355, 0.6324,0.3082, 0.0873));
  rects.push_back(Rect(0.1270,0.9689, 0.8147,1.0000, 0.0214));
  rects.push_back(Rect(0.0000,0.9689, 0.1270,1.0000, 0.0040));

  rects.push_back(Rect(0.9134,0.1355, 1.0000,0.2210, 0.0074));
  rects.push_back(Rect(0.9134,0.2210, 1.0000,0.8350, 0.0532));
  rects.push_back(Rect(0.9058,0.2210, 0.9134,0.8350, 0.0047));


  rects.push_back(Rect(0.6324,0.1355, 0.8147,0.3082, 0.0315));
  rects.push_back(Rect(0.6324,0.3082, 0.8147,0.9689, 0.1205));
  rects.push_back(Rect(0.1270,0.3082, 0.6324,0.9689, 0.3339));

  int b=13;
  int nearBox_count = 0;
  //double area=0;
  double TotalArea=0;

  for (int x = 0; x < rects.size(); ++x) {

    if (rects[b].NearestBox(rects[x])) {  
      if (x==b) {
        continue; //this is our box , so do not count it.
      }

    nearBox_count++;
    printf("box[%d] is nearest to box[%d] and has area %f \n", (b+1), (x+1), rects[x].area);

    TotalArea +=rects[x].area;

    }
  }

  printf("Total number of nearest box for box[%d] = %d, and the sum of area is= %f \n", (b+1), nearBox_count, TotalArea );

  return 0;
}

It prints the results-

box[14] is nearest to box[1] and has area 0.000000 
box[14] is nearest to box[3] and has area 0.000000 
box[14] is nearest to box[8] and has area 0.000000 
box[14] is nearest to box[15] and has area 0.000000 
Total number of nearest box for box[14] = 4, and the sum of area is= 0.000000 

So by comparing the results with the picture above, you can see that it returns the value of the nearest boxes but fails to return the value of their corresponding areas.

Can anyone help me to fix that?

like image 286
aries0152 Avatar asked Aug 05 '13 16:08

aries0152


1 Answers

You're not actually using the area in rect[x]. You don't need the extra area variable, either. Just do something like:

printf("box[%d] is nearest to box[%d] and has area %f \n", (b+1), (x+1), rects[x].area);

TotalArea += rects[x].area;

Also, from masad's comment, you don't assign area in the constructor. You need to add a line:

area = Area;
like image 164
Geobits Avatar answered Sep 20 '22 02:09

Geobits