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Problem solution · Java

CCC 2006 S3 - Tin Can Telephone

CCC 2006 S3 - Tin Can Telephone: 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
139 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 2006 S3 - Tin Can Telephone, 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

  • 139 lines of Java from the credited upstream file ccc06s3.java.
  • The implementation visibly relies on sequence storage.
  • 2 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 2006 S3 - Tin Can Telephone · JavaJava
Use this to learn the idea, then write your own version.
// CCC2006s3TinCanTelephone//// this is a geometry problem which can be simplified to:// are two line segments touching or crossing?// the answer is yes if their point of intersection is on// both lines.//// you need to consider 4 cases in total// parallel//    both vertical//    both angled// non-parallel//    one vertical//    both angled// import java.awt.*;import hsa.*; public class CCC2006s3TinCanTelephone{    static Console c;     public static void main (String[] args)    {	c = new Console ();	TextInputFile f = new TextInputFile ("s3.1.in");	int xr, yr, xj, yj;	int n, corners;	int x0, y0, x1, y1, x2, y2;	boolean touching;	int touch = 0; 	xr = f.readInt ();	yr = f.readInt ();	xj = f.readInt ();	yj = f.readInt (); 	n = f.readInt ();	for (int i = 0 ; i < n ; i++)	{	    touching = false;	    corners = f.readInt ();	    x0 = f.readInt ();	    y0 = f.readInt ();	    x1 = x0;	    y1 = y0;	    for (int j = 1 ; j < corners ; j++)	    {		x2 = f.readInt ();		y2 = f.readInt ();		touching = touching || touchingSegments (xr, yr, xj, yj, x1, y1, x2, y2);		x1 = x2;		y1 = y2;	    }	    touching = touching || touchingSegments (xr, yr, xj, yj, x1, y1, x0, y0);	    if (touching)		touch++;	}	c.println (touch);    }      public static boolean touchingSegments (int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)    {	double m1, m2, b1, b2, xi, yi;	if (x1 == x2)	{	    m1 = Double.MAX_VALUE;	    b1 = 0;	}	else	{	    m1 = (y1 - y2) / (x1 - x2);	    b1 = -m1 * x2 + y2;	}	if (x3 == x4)	{	    m2 = Double.MAX_VALUE;	    b2 = 0;	}	else	{	    m2 = (y3 - y4) / (x3 - x4);	    b2 = -m2 * x4 + y4;	} 	// segments are parallel	if (m1 == m2)	{	    // angled lines	    // they touch if their y-intercepts are the same and	    // one point of one segment is within the other segment	    if (m1 != Double.MAX_VALUE && m2 != Double.MAX_VALUE)		return b1 == b2 && (between (x3, x1, x2) || between (x4, x1, x2)); 	    // lines are both vertical	    // they touch if their x's are the same and y's overlap	    else		return x1 == x3 && (between (y3, y1, y2) || between (y4, y1, y2));	}	else	{	    // normal angled lines,	    // touch if point of intersection is on both lines	    if (m1 != Double.MAX_VALUE && m2 != Double.MAX_VALUE)	    {		xi = (b2 - b1) / (m1 - m2);		yi = m1 * xi + b1;		return between (xi, x1, x2) && between (yi, y1, y2) && between (xi, x3, x4) && between (yi, y3, y4);	    } 	    // one line is a vertical	    // touch if point of intersection is on both lines	    else	    {		if (m1 == Double.MAX_VALUE)		{		    xi = x1;		    yi = m2 * xi + b2;		}		else		{		    xi = x3;		    yi = m1 * xi + b1;		}		return between (xi, x1, x2) && between (yi, y1, y2) && between (xi, x3, x4) && between (yi, y3, y4);	    }	}    }      // returns true if x is between a and b    public static boolean between (double x, int a, int b)    {	return (x >= a && x <= b) || (x <= a && x >= b);    }}

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