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Kleejev algoritem (dolžina spoja segmentov črte)

Glede na začetni in končni položaj segmentov na premici je naloga vzeti zvezo vseh danih segmentov in poiskati dolžino, ki jo pokrivajo ti segmenti.
Primeri:   

  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)

Pristop:



java znak v niz

Algoritem je predlagal Klee leta 1977. Časovna kompleksnost algoritma je O (N log N). Dokazano je, da je ta algoritem najhitrejši (asimptotično) in tega problema ni mogoče rešiti z boljšo kompleksnostjo. 

Opis:  
1) Postavite vse koordinate vseh segmentov v pomožni niz točk []. 
2) Razvrsti po vrednosti koordinat. 
3) Dodaten pogoj za razvrščanje - če so koordinate enake, namesto desne vstavite tisto, ki je leva koordinata poljubnega segmenta. 
4) Zdaj pojdite skozi celotno matriko s števcem 'štetja' prekrivajočih se segmentov. 
5) Če je štetje večje od nič, se rezultat prišteje k razliki med točkama [i] - točkama [i-1]. 
6) Če trenutni element pripada levemu koncu, povečamo 'count', sicer ga zmanjšamo.
Ilustracija:  

odstranite zadnji znak iz niza
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) 

Izhod
9

Časovna zapletenost: O(n * log n)
Pomožni prostor: O(n)