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Antti H S Laaksonen 2017-04-22 15:19:48 +03:00
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chapter30.tex
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@ -5,13 +5,13 @@
Many geometric problems can be solved using
\key{sweep line} algorithms.
The idea in such algorithms is to represent
the problem as a set of events that correspond
an instance of the problem as a set of events that correspond
to points in the plane.
The events are processed in increasing order
according to their x or y coordinate.
As an example, let us consider a problem
where there is a company that has $n$ employees,
As an example, let us consider the following problem:
There is a company that has $n$ employees,
and we know for each employee their arrival and
leaving times on a certain day.
Our task is to calculate the maximum number of
@ -162,12 +162,15 @@ there are three intersection points:
\end{center}
It is easy to solve the problem in $O(n^2)$ time,
because we can go through all possible pairs of segments
because we can go through all possible pairs of line segments
and check if they intersect.
However, we can solve the problem more efficiently
in $O(n \log n)$ time using a sweep line algorithm.
in $O(n \log n)$ time using a sweep line algorithm
and a range query data structure.
The idea is to generate three types of events:
The idea is to process the endpoints of the line
segments from left to right and
focus on three types of events:
\begin{enumerate}[noitemsep]
\item[(1)] horizontal segment begins
\item[(2)] horizontal segment ends
@ -209,11 +212,10 @@ horizontal segments whose y coordinate is between
$y_1$ and $y_2$, and add this number to the total
number of intersection points.
An appropriate data structure for storing
y coordinates of horizontal segments is either
a binary indexed tree or a segment tree,
To store y coordinates of horizontal segments,
we can use a binary indexed or a segment tree,
possibly with index compression.
Using such structures, processing each event
Using such a structure, processing each event
takes $O(\log n)$ time, so the total running
time of the algorithm is $O(n \log n)$.
@ -222,7 +224,7 @@ time of the algorithm is $O(n \log n)$.
\index{closest pair}
Given a set of $n$ points, our next problem is
to find two points whose distance is minimum.
to find two points whose Euclidean distance is minimum.
For example, if the points are
\begin{center}
\begin{tikzpicture}[scale=0.7]
@ -412,7 +414,7 @@ add each point to the hull.
Always after adding a point to the hull,
we make sure that the last line segment
in the hull does not turn left.
As long as this holds, we repeatedly remove the
As long as this does not hold, we repeatedly remove the
second last point from the hull.
The following pictures show how