Algoritmo de Klee (longitud de unión de segmentos de una línea)

Dadas las posiciones inicial y final de los segmentos en una línea, la tarea es tomar la unión de todos los segmentos dados y encontrar la longitud cubierta por estos segmentos.
Ejemplos:  

Input : segments[] = {{2, 5}, {4, 8}, {9, 12}}
Output : 9 
Explanation:
segment 1 = {2, 5}
segment 2 = {4, 8}
segment 3 = {9, 12}
If we take the union of all the line segments,
we cover distances [2, 8] and [9, 12]. Sum of 
these two distances is 9 (6 + 3)

Acercarse:

El algoritmo fue propuesto por Klee en 1977. La complejidad temporal del algoritmo es O (N log N). Se ha comprobado que este algoritmo es el más rápido (asintóticamente) y este problema no se puede resolver con una mayor complejidad. 

Descripción: 
1) Poner todas las coordenadas de todos los segmentos en una array auxiliar puntos[]. 
2) Ordenarlo según el valor de las coordenadas. 
3) Una condición adicional para la clasificación: si hay coordenadas iguales, inserte la que sea la coordenada izquierda de cualquier segmento en lugar de la derecha. 
4) Ahora recorra toda la array, con el contador «recuento» de segmentos superpuestos. 
5) Si el conteo es mayor que cero, entonces el resultado se suma a la diferencia entre los puntos[i] – puntos[i-1]. 
6) Si el elemento actual pertenece al extremo izquierdo, aumentamos «contar», de lo contrario lo reducimos.
Ilustración: 

Lets take the example :
segment 1 : (2,5)
segment 2 : (4,8)
segment 3 : (9,12)

Counter = result = 0;
n = number of segments = 3;

for i=0,  points[0] = {2, false}
          points[1] = {5, true}
for i=1,  points[2] = {4, false}
          points[3] = {8, true}
for i=2,  points[4] = {9, false}
          points[5] = {12, true}

Therefore :
points = {2, 5, 4, 8, 9, 12}
         {f, t, f, t, f, t}

after applying sorting :
points = {2, 4, 5, 8, 9, 12}
         {f, f, t, t, f, t}

Now,
for i=0, result = 0;
         Counter = 1;

for i=1, result = 2;
         Counter = 2;

for i=2, result = 3;
         Counter = 1;

for i=3, result = 6;
         Counter = 0;

for i=4, result = 6;
         Counter = 1;

for i=5, result = 9;
         Counter = 0;

Final answer = 9;

// C++ program to implement Klee's algorithm
#include<bits/stdc++.h>
using namespace std;
 
// Returns sum of lengths covered by union of given
// segments
int segmentUnionLength(const vector<
                          pair <int,int> > &seg)
{
    int n = seg.size();
 
    // Create a vector to store starting and ending
    // points
    vector <pair <int, bool> > points(n * 2);
    for (int i = 0; i < n; i++)
    {
        points[i*2]     = make_pair(seg[i].first, false);
        points[i*2 + 1] = make_pair(seg[i].second, true);
    }
 
    // Sorting all points by point value
    sort(points.begin(), points.end());
 
    int result = 0; // Initialize result
 
    // To keep track of counts of
    // current open segments
    // (Starting point is processed,
    // but ending point
    // is not)
    int Counter = 0;
 
    // Traverse through all points
    for (unsigned i=0; i<n*2; i++)
    {
        // If there are open points, then we add the
        // difference between previous and current point.
        // This is interesting as we don't check whether
        // current point is opening or closing,
        if (Counter)
            result += (points[i].first -
                        points[i-1].first);
 
        // If this is an ending point, reduce, count of
        // open points.
        (points[i].second)? Counter-- : Counter++;
    }
    return result;
}
 
// Driver program for the above code
int main()
{
    vector< pair <int,int> > segments;
    segments.push_back(make_pair(2, 5));
    segments.push_back(make_pair(4, 8));
    segments.push_back(make_pair(9, 12));
    cout << segmentUnionLength(segments) << endl;
    return 0;
}

// Java program to implement Klee's algorithm
import java.io.*;
import java.util.*;
 
class GFG {
 
  // to use create a pair of segments
  static class SegmentPair
  {
    int x,y;
    SegmentPair(int xx, int yy){
      this.x = xx;
      this.y = yy;
    }
  }
 
  //to create a pair of points
  static class PointPair{
    int x;
    boolean isEnding;
    PointPair(int xx, boolean end){
      this.x = xx;
      this.isEnding = end;
    }
  }
 
  // creates the comparator for comparing objects of PointPair class
  static class Comp implements Comparator<PointPair>
  {
     
    // override the compare() method
    public int compare(PointPair p1, PointPair p2)
    {
      if (p1.x < p2.x) {
        return -1;
      }
      else {
        if(p1.x == p2.x){
          return 0;
        }else{
          return 1;
        }
      }
    }
  }
 
  public static int segmentUnionLength(List<SegmentPair> segments){
    int n = segments.size();
 
    // Create a list to store
    // starting and ending points
    List<PointPair> points = new ArrayList<>();
    for(int i = 0; i < n; i++){
      points.add(new PointPair(segments.get(i).x,false));
      points.add(new PointPair(segments.get(i).y,true));
    }
     
    // Sorting all points by point value
    Collections.sort(points, new Comp());
 
    int result = 0; // Initialize result
 
    // To keep track of counts of
    // current open segments
    // (Starting point is processed,
    // but ending point
    // is not)
    int Counter = 0;
 
    // Traverse through all points
    for(int i = 0; i < 2 * n; i++)
    {
       
      // If there are open points, then we add the
      // difference between previous and current point.
      // This is interesting as we don't check whether
      // current point is opening or closing,
      if (Counter != 0)
      {
        result += (points.get(i).x - points.get(i-1).x);
      }
 
      // If this is an ending point, reduce, count of
      // open points.
      if(points.get(i).isEnding)
      {
        Counter--;
      }
      else
      {
        Counter++;
      }
    }
    return result;
  }
 
  // Driver Code
  public static void main (String[] args) {
    List<SegmentPair> segments = new ArrayList<>();
    segments.add(new SegmentPair(2,5));
    segments.add(new SegmentPair(4,8));
    segments.add(new SegmentPair(9,12));
    System.out.println(segmentUnionLength(segments));
  }
}
 
// This code is contributed by shruti456rawal

# Python program for the above approach
 
def segmentUnionLength(segments):
   
    # Size of given segments list
    n = len(segments)
     
    # Initialize empty points container
    points = [None] * (n * 2)
     
    # Create a vector to store starting
    # and ending points
    for i in range(n):
        points[i * 2] = (segments[i][0], False)
        points[i * 2 + 1] = (segments[i][1], True)
         
    # Sorting all points by point value
    points = sorted(points, key=lambda x: x[0])
     
    # Initialize result as 0
    result = 0
     
    # To keep track of counts of current open segments
    # (Starting point is processed, but ending point
    # is not)
    Counter = 0
     
    # Traverse through all points
    for i in range(0, n * 2):
       
        # If there are open points, then we add the
        # difference between previous and current point.
        if (i > 0) & (points[i][0] > points[i - 1][0]) &  (Counter > 0):
            result += (points[i][0] - points[i - 1][0])
             
        # If this is an ending point, reduce, count of
        # open points.
        if points[i][1]:
            Counter -= 1
        else:
            Counter += 1
    return result
 
 
# Driver code
if __name__ == '__main__':
    segments = [(2, 5), (4, 8), (9, 12)]
    print(segmentUnionLength(segments))

// JavaScript program to implement Klee's algorithm
 
// Returns sum of lengths covered by union of given
// segments
function segmentUnionLength(seg)
{
    let n = seg.length;
 
    // Create a vector to store starting and ending
    // points
    let points = new Array(2*n);
    for (let i = 0; i < n; i++)
    {
        points[i*2] = [seg[i][0], false];
        points[i*2 + 1] = [seg[i][1], true];
    }
 
    // Sorting all points by point value
    points.sort(function(a, b){
        return a[0] - b[0];
    });
     
    let result = 0; // Initialize result
 
    // To keep track of counts of
    // current open segments
    // (Starting point is processed,
    // but ending point
    // is not)
    let Counter = 0;
 
    // Traverse through all points
    for (let i=0; i<n*2; i++)
    {
        // If there are open points, then we add the
        // difference between previous and current point.
        // This is interesting as we don't check whether
        // current point is opening or closing,
        if (Counter)
            result += (points[i][0] - points[i-1][0]);
 
        // If this is an ending point, reduce, count of
        // open points.
        if(points[i][1]){
            Counter = Counter - 1;
        }
        else{
            Counter = Counter + 1;
        }
    }
    return result;
}
 
let segments = new Array();
segments.push([2, 5]);
segments.push([4, 8]);
segments.push([9, 12]);
console.log(segmentUnionLength(segments));
 
// The code is contributed by Gautam goel (gautamgoel962)

9

Complejidad de tiempo: O(n * log n)
Espacio auxiliar: O(n)

Este artículo es una contribución de Aarti_Rathi y Abhinandan Mittal . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.