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Klees Algorithmus (Länge der Vereinigung von Segmenten einer Linie)

Bei gegebenen Anfangs- und Endpositionen von Segmenten auf einer Linie besteht die Aufgabe darin, die Vereinigung aller gegebenen Segmente zu bilden und die von diesen Segmenten abgedeckte Länge zu ermitteln.
Beispiele:   

  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)

Ansatz:



Der Algorithmus wurde 1977 von Klee vorgeschlagen. Die zeitliche Komplexität des Algorithmus beträgt O (N log N). Es ist erwiesen, dass dieser Algorithmus der schnellste (asymptotisch) ist und dieses Problem nicht mit einer höheren Komplexität gelöst werden kann. 

Beschreibung :  
1) Geben Sie alle Koordinaten aller Segmente in ein Hilfsarray „points[] ein. 
2) Sortieren Sie es nach dem Wert der Koordinaten. 
3) Eine zusätzliche Bedingung für die Sortierung: Wenn es gleiche Koordinaten gibt, fügen Sie diejenige ein, die die linke Koordinate eines beliebigen Segments ist, anstelle einer rechten. 
4) Gehen Sie nun das gesamte Array mit dem Zähler „Anzahl“ der überlappenden Segmente durch. 
5) Wenn die Anzahl größer als Null ist, wird das Ergebnis zur Differenz zwischen Punkten[i] und Punkten[i-1] addiert. 
6) Wenn das aktuelle Element zum linken Ende gehört, erhöhen wir „count“, andernfalls verringern wir es.
Illustration:  

Lets take the example : segment 1 : (25) segment 2 : (48) segment 3 : (912) 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++
// C++ program to implement Klee's algorithm #include   using namespace std; // Returns sum of lengths covered by union of given // segments int segmentUnionLength(const vector<   pair <intint> > &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 <intint> > 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
// 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 xy;  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).xfalse));  points.add(new PointPair(segments.get(i).ytrue));  }    // 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(25));  segments.add(new SegmentPair(48));  segments.add(new SegmentPair(912));  System.out.println(segmentUnionLength(segments));  } } // This code is contributed by shruti456rawal 
Python3
# 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)) 
C#
using System; using System.Collections; using System.Collections.Generic; using System.Linq; // C# program to implement Klee's algorithm class HelloWorld {  class GFG : IComparer<KeyValuePair<int bool>>  {  public int Compare(KeyValuePair<int bool> xKeyValuePair<int bool> y)  {  // CompareTo() method  return x.Key.CompareTo(y.Key);  }  }  // Returns sum of lengths covered by union of given  // segments  public static int segmentUnionLength(List<KeyValuePair<intint>> seg)  {  int n = seg.Count;  // Create a vector to store starting and ending  // points  List<KeyValuePair<int bool>> points = new List<KeyValuePair<int bool>>();  for(int i = 0; i < 2*n; i++){  points.Add(new KeyValuePair<int bool> (0true));  }   for (int i = 0; i < n; i++)  {  points[i*2] = new KeyValuePair<int bool> (seg[i].Key false);  points[i*2 + 1] = new KeyValuePair<int bool> (seg[i].Value true);  }  // Sorting all points by point value  GFG gg = new GFG();  points.Sort(gg);  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<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 != 0)  result += (points[i].Key - points[i-1].Key);  // If this is an ending point reduce count of  // open points.  if(points[i].Value != false){  Counter--;  }  else{  Counter++;  }  }  return result;  }  static void Main() {  List<KeyValuePair<intint>> segments = new List<KeyValuePair<intint>> ();  segments.Add(new KeyValuePair<intint> (2 5));  segments.Add(new KeyValuePair<intint> (4 8));  segments.Add(new KeyValuePair<intint> (9 12));  Console.WriteLine(segmentUnionLength(segments));  } } // The code is contributed by Nidhi goel.  
JavaScript
// 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) 

Ausgabe
9

Zeitkomplexität: O(n * log n)
Hilfsraum: An)