Write a test program that prompts the user to enter the set size and the points and displays the points that form a convex hull. Note that when you debug the code, you will discover that the algorithm overlooked two cases (1) when t1 = t0 and (2) when there is a point that is on the same line from t0 to t1. When either case happens, replace t1 by point p if the distance from t0 to p is greater than the distance from t0 to t1. Here is a sample run:

I keep getting event overload errors, thanks in advance.

#include

using namespace std;

struct Point

{ //2D Convex Plot int x;

int y;

}; // 0 --> p, q and r are colinear (mid.point)

// 1 --> Clockwise

// 2 --> Counterclockwise

int orientation(Point p, Point q, Point r) //checks point for relation

{

int val = (q.y - p.y) * (r.x - q.x) - (q.x - p.x) * (r.y - q.y);

if (val == 0)

return 0; // colinear

return (val > 0) ? 1 : 2; // clock or counterclock wise

}

// Prints convex hull of a set of n points.

void convexHull(Point points[])

{ const int n = 6;

//Let t1 and t2 be the top first and second element in stack H; //if (pi is on the left side of the direct line from t2 to t1)

if (n

return;

// Initialize Result

int next[6];

for (int i = 0; i

next[i] = -1;

// Find the leftmost point

int l = 0;

for (int i = 1; i

if (points[i].x

l = i;

// Start from leftmost point, keep moving counterclockwise

// until reach the start point again

int p = l, q;

do

{

// Search for a point 'q' such that orientation(p, i, q) is

// counterclockwise for all points 'i'

q = (p + 1) % n;

for (int i = 0; i

if (orientation(points[p], points[i], points[q]) == 2)

q = i;

next[p] = q; // Add q to result as a next point of p

p = q; // Set p as q for next iteration

}

while (p != l);

// Print Result

for (int i = 0; i

{

if (next[i] != -1)

cout

}

}

int main()

{

Point points[] = { { 1, 2.4 }, { 2.5, 2 }, { 1.5, 34.5 }, { 5.5, 6 }, { 6, 2.4 },

{ 5.5, 9 }};

cout

int n = sizeof(points) / sizeof(points[0]);

convexHull(points, n);

return 0;

};

18: Developing Efficient Algorithms ho Repeat Step 2: A convex hullis expanded repeatedly (a) ho is the rightmost lowest point in S. (b) Step 2 finds point t. (c) A convex hull is expanded repeatedly. (d) A convex hull is found when t, becomes ho Sample Run for Exercise 18_11 Enter input data for the program (Sample data provided below. You may modify it.) 1 2.4 2.5 2 1.5 34.5 5.5 6 6 2.4 5.5 9 Show the Sample Output Using the Preceeding Input Reset command>Exercise18 11 How many points are in the set? 6 Enter 6 points: 1 2.4 2.5 2 1.5 34.5 5.5 6 6 2.4 5.59 The convex hull is (2.5, 2.0) (6.0, 2.4) (5.5, 9.0) (1.5, 34.5) (1.0, 2.4) command> 18: Developing Efficient Algorithms ho Repeat Step 2: A convex hullis expanded repeatedly (a) ho is the rightmost lowest point in S. (b) Step 2 finds point t. (c) A convex hull is expanded repeatedly. (d) A convex hull is found when t, becomes ho Sample Run for Exercise 18_11 Enter input data for the program (Sample data provided below. You may modify it.) 1 2.4 2.5 2 1.5 34.5 5.5 6 6 2.4 5.5 9 Show the Sample Output Using the Preceeding Input Reset command>Exercise18 11 How many points are in the set? 6 Enter 6 points: 1 2.4 2.5 2 1.5 34.5 5.5 6 6 2.4 5.59 The convex hull is (2.5, 2.0) (6.0, 2.4) (5.5, 9.0) (1.5, 34.5) (1.0, 2.4) command>