Merge pull request #219 from neha030/dev0

Klee's Algorithm added

Author: smv1999

Dynamic Programming/Klee's Algorithm.cpp

@@ -0,0 +1,103 @@
+/*    
+    Description: Given a Point with x and y coordinates,this algorithm finds the union of the
+                 lengths covered by all these line segments.
+    This algorithm was proposed by Klee in 1977 and is asymptotically the fastest.     */
+
+
+#include<bits/stdc++.h>
+#define x first
+#define y second
+
+using namespace std;
+
+//Structure to store the point coordinates.
+typedef struct points {
+	int x, y;
+}points;
+
+//vector to indicate the points in the plane.
+vector<points>plane;
+
+//function to find the union of the lengths.
+int find_union(vector<points>line) {
+
+	int size = line.size();
+
+	//This vector first stores the coordinate and then the bool value.
+	//Here,the starting coordinate is marked false and the ending coordinate as true in pair.
+	vector<pair<int, bool>> points(size * 2);
+
+	for (int i = 0; i < size; i++)
+	{
+
+		points[i * 2] = make_pair(line[i].x, false);
+		points[i * 2 + 1] = make_pair(line[i].y, true);
+
+	}
+
+	//This is to sort all the ending points.
+	sort(points.begin(), points.end());
+
+	//Initializing variable answer which will store the final length and counter which will keep track of the opening and closing segment.
+	int answer = 0, counter = 0;
+
+	//traversing through all the ending points
+	for (int i = 0; i < size * 2; i++)
+	{
+		//for adding the difference of current and previous points to answer.
+		if (counter)
+			answer += (points[i].x - points[i - 1].x);
+
+		//for the endpoint of the segment, decrements the counter or else increment the counter.
+		if (points[i].y)
+			counter--;
+		else
+			counter++;
+	}
+
+	return answer;
+}
+
+int main()
+{
+
+	//declaration of the number of points.
+	int n;
+
+	cout << "Enter the desired number of points:\n";
+	cin >> n;
+	cout << "Enter the x and y coordinates for the points with a space:\n";
+
+	plane.resize(n);
+
+	//For input of x and y coordinates of the desired points
+	for (int i = 0; i < n; i++)
+	{
+		cin >> plane[i].x >> plane[i].y;
+	}
+
+	//Outputs the final length of the union the length of segments.
+	cout << "The union of the length of the line segments is :\n";
+
+	int ans = find_union(plane);
+
+	cout << ans;
+	return 0;
+
+}
+
+/*    Time Complexity: O(N log N)
+      Space Complexity: O(N)        */
+
+/* Test Case:
+*
+Sample Input: Enter the desired number of points:
+			  4
+			  Enter the x and y coordinates for the points with a space:
+			  2 4
+			  5 6
+			  3 7
+			  1 6
+Sample Output: The union of the length of the line segments is:
+			  6
+			  */