CPMagical CP Library

Problem solution · Java

CCC 2000 S5 - Sheep and Coyotes

CCC 2000 S5 - Sheep and Coyotes: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
89 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2000 S5 - Sheep and Coyotes, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 89 lines of Java from the credited upstream file ccc00s5.java.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2000 S5 - Sheep and Coyotes · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2000// Problem S5 Sheep and Coyotes // This is an exhaustive search problem// given the location of up to 100 sheep// find the closest to each distance from 0 to 1000 in steps of 0.01// (y = 0 in all cases: they are the coyotes!)// eliminating dups. // file consists of number of sheep and their loactions.// output is a list of closest sheep import java.awt.*;import hsa.*;  public class S5SheepandCoyotes{    static Console c;     public static void main (String [] args)    {	c = new Console ();	Point sheep [] = new Point [100];	int eaten [] = new int [100]; // list of possibly eaten sheep	int n, ne, s;	double d, m;	boolean dup; 	TextInputFile fi = new TextInputFile ("sheep.in5");	TextOutputFile fo = new TextOutputFile ("sheep.ou5"); 	n = fi.readInt ();	for (int i = 0 ; i < n ; i++)	    sheep [i] = new Point (fi.readDouble (), fi.readDouble ());	ne = 0;	for (double x = 0 ; x <= 1000.0 ; x = x + 0.01)	{ 	    // calculate distance to closest sheep	    s = 0;	    m = 9999999.0;	    for (int i = 0 ; i < n ; i++)	    {		d = distance (sheep [i], x);		if (d < m)		{		    m = d;		    s = i;		}	    } 	    // check if that sheep is already on the list	    dup = false;	    for (int j = 0 ; j < ne ; j++)		if (s == eaten [j])		    dup = true; 	    // if not on the list, add to the list	    if (!dup)		eaten [ne++] = s;	}	for (int j = 0 ; j < ne ; j++)	{	    fo.println ("The sheep at (" + sheep [eaten [j]].x + ", " + sheep [eaten [j]].y + ") might be eaten.");	    c.println ("The sheep at (" + sheep [eaten [j]].x + ", " + sheep [eaten [j]].y + ") might be eaten.");	}	fi.close ();	fo.close ();    }      public static double distance (Point p, double x)    {	return (p.x - x) * (p.x - x) + p.y * p.y;    }} class Point{    public double x, y;    public Point (double a, double b)    {	x = a;	y = b;    }}

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗