Finding the number of triangles amongst horizontal and vertical line segments - GeeksforGeeks

Finding the number of triangles amongst horizontal and vertical line segments - GeeksforGeeks
Given 'n' line segments, each of them is either horizontal or vertical, find the maximum number of triangles(including triangles with zero area) that can be formed by joining the intersection points of the line segments.
No two horizontal line segments overlap, nor do two vertical line segments. A line is represented using two points(four integers, first two being the x and y coordinates, respectively for the first point and the other two being the x and y coordinates for the second point)

The idea is based on Sweep Line Algorithm.

Building a solution in steps:

1. Store both points of all line segments with the corresponding event(described below) in a vector and sort all the points, in non-decreasing order of their x coordinates.
2. Let’s now imagine a vertical line that we sweep across all these points and describe 3 events, based on which point we currently are:
   • in – leftmost point of a horizontal line segment
   • out – rightmost point of a horizontal line segment
   • a vertical line
3. We call the region “active” or the horizontal lines “active” that have had the first event but not second. We will have a BIT(Binary indexed tree) to store the ‘y’ coordinates of all active lines.
4. Once a line becomes inactive, we remove its ‘y’ from the BIT.
5. When an event of third type occurs, that is, when we are at a vertical line, we query the tree in range of its ‘y’ coordinates and add the result to the number of intersection points so far.
6. Finally, we will have the number of points of intersections, say m, then the number of triangles (including zero area) will be mC3.

Time Complexity: O( n * log(n) + n * log(maximum_y) )

struct point

{

    int x, y;

    point(int a, int b)

    {

        x = a, y = b;

    }

};

 

// Note: Global arrays are initially zero

// array to store BIT and vector to store

// the points and their corresponding event number,

// in the second field of the pair

int bit[maxy];

vector<pair<point, int> > events;

 

// compare function to sort in order of non-decreasing

// x coordinate and if x coordinates are same then

// order on the basis of events on the points

bool cmp(pair<point, int> &a, pair<point, int> &b)

{

    if ( a.first.x != b.first.x )

        return a.first.x < b.first.x;

 

    //if the x coordinates are same

    else

    {

        // both points are of the same vertical line

        if (a.second == 3 && b.second == 3)

        {

            return true;

        }

 

        // if an 'in' event occurs before 'vertical'

        // line event for the same x coordinate

        else if (a.second == 1 && b.second == 3)

        {

            return true;

        }

 

        // if a 'vertical' line comes before an 'in'

        // event for the same x coordinate, swap them

        else if (a.second == 3 && b.second == 1)

        {

            return false;

        }

 

        // if an 'out' event occurs before a 'vertical'

        // line event for the same x coordinate, swap.

        else if (a.second == 2 && b.second == 3)

        {

            return false;

        }

 

        //in all other situations

        return true;

    }

}

 

// update(y, 1) inserts a horizontal line at y coordinate

// in an active region, while update(y, -1) removes it

void update(int idx, int val)

{

    while (idx < maxn)

    {

        bit[idx] += val;

        idx += idx & (-idx);

    }

}

 

// returns the number of lines in active region whose y

// coordinate is between 1 and idx

int query(int idx)

{

    int res = 0;

    while (idx > 0)

    {

        res += bit[idx];

        idx -= idx & (-idx);

    }

    return res;

}

 

// inserts a line segment

void insertLine(point a, point b)

{

    // if it is a horizontal line

    if (a.y == b.y)

    {

        int beg = min(a.x, b.x);

        int end = max(a.x, b.x);

 

        // the second field in the pair is the event number

        events.push_back(make_pair(point(beg, a.y), 1));

        events.push_back(make_pair(point(end, a.y), 2));

    }

 

    //if it is a vertical line

    else

    {

        int up = max(b.y, a.y);

        int low = min(b.y, a.y);

 

        //the second field of the pair is the event number

        events.push_back(make_pair(point(a.x, up), 3));

        events.push_back(make_pair(point(a.x, low), 3));

    }

}

 

// returns the number of intersection points between all

// the lines, vertical and horizontal, to be run after the

// points have been sorted using the cmp() function

int findIntersectionPoints()

{

    int intersection_pts = 0;

    for (int i = 0 ; i < events.size() ; i++)

    {

        //if the current point is on an 'in' event

        if (events[i].second == 1)

        {

            //insert the 'y' coordinate in the active region

            update(events[i].first.y, 1);

        }

 

        // if current point is on an 'out' event

        else if (events[i].second == 2)

        {

            // remove the 'y' coordinate from the active region

            update(events[i].first.y, -1);

        }

 

        // if the current point is on a 'vertical' line

        else

        {

            // find the range to be queried

            int low = events[i++].first.y;

            int up = events[i].first.y;

            intersection_pts += query(up) - query(low);

        }

    }

    return intersection_pts;

}

 

// returns (intersection_pts)C3

int findNumberOfTriangles()

{

    int pts = findIntersectionPoints();

    if ( pts >= 3 )

        return ( pts * (pts - 1) * (pts - 2) ) / 6;

    else

        return 0;

}

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