Assignment 08 Solution

A set of lines in the plane partitions the points in the plane into regions, 
such that two points are in the same region iff the line segment joining
them does not intersect any of the given lines. A point contained in any
of the given lines does not belong to any region.

Given a set of n lines in the plane, and a set of m points, you have to find
the number of regions formed by the n lines and the number of the given points 
contained in each region. 

A binary tree can be used to represent the set of lines. Every
node in the tree contains a line, and every empty subtree represents a region.
Every node in the tree represents some region formed by a subset of the lines.
The tree is built by inserting one line at a time. Initially the empty tree
represents the whole plane. When the first line is inserted, it is placed in
the root node, the left subtree is the region to the left of the line and
the right subtree is the region to the right. In general, when inserting a
new line, some line segment (possibly infinite) contained in that line will be 
recursively inserted in some of the subtrees of the tree.

When some segment is to be inserted in a subtree containing a line l in the
root, check if l intersects the segment. If so, insert the part of the segment
to the left of l in the left subtree and the other part in the right. If the
segment is completely to the left or right of l, insert it in only that
subtree. When an empty tree is reached, a new node is created containing the
line being inserted. This corresponds to an original region being split
into two by the new line. If the lines are all parallel, this is equivalent
to insertion in ordinary binary search trees. In general, there may be 
many nodes containing the same line.

It can be shown that if the lines are inserted in random order, the average
depth of a region is O(log n). The depth of a region is the number of nodes
whose subtree contains the region. The total number of nodes in the tree can 
be O(n^2) since the number of regions can be as large as this. After building 
the tree, locating the region containing a given point can be done in time 
proportional to the  depth of the region. 

There is a better way of inserting if we also store the line segments that
form the boundary of each region. Then after finding one region to be split
by the new line, other regions can be found by simply intersecting the
new line with the boundary of the current region. This takes total O(n) time
for insertion and is also needed for deletion. You don't have to do this
for the lab. Alternatively, we can find the intersection points of the new 
line with all earlier lines, sort these, locate the regions containing the 
midpoints of the segments formed, and split those. Note that this gives a 
simple way of computing the number of regions formed.


Input Format
The first line of input specifies n and m the number of lines and points,
1 <= n <= 5000, 1 <= m <= 1000000. The next n lines specify the equations
of the n lines. Each line contains 3 coefficients a, b, c where the equation
of the line is ax+by = c.  The next m lines give the coordinates of
the points.  It can be assumed that each of the numbers is a floating point 
number with at most 2 decimal places and at most 10000 in magnitude.

Output Format
The first line should print out R, the total number of regions formed,
and the height H of the binary tree, if the lines are inserted in the
order given.  Then print out the number of regions containing a given 
number of the points, in increasing order of number of points. Print two 
numbers per line, the number of points in the region followed by the number
of regions containing those many points. This should be printed only
if there is at least one region with those many points.

Sample Input        Output
4 8                 9 4
0 1 0               0 5
0 1 2               1 3
1 0 0               3 1
1 0 2
1 1
0 0
1 3
3 3
2 4
1 5
-1 -1
1 4

Submit a single file RollNo_8.cpp
There will be NO extension of time. Submission time is 5-00 p.m. and 
the link will close at 5-05 p.m.