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Problem solution · Java

CCC 2000 S3 - Surfing

CCC 2000 S3 - Surfing: 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
133 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 S3 - Surfing, 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

  • 133 lines of Java from the credited upstream file ccc00s3.java.
  • The implementation visibly relies on sequence storage.
  • 12 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 S3 - Surfing · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2000// Problem S3J5: Surfin'//// This reads a series of web pages and the idea is to find the links// to other pages and to determine if you can "get there from here"// Warshall's Transitive closure and 2D adjacency matrix are used.//// Input file consists of the the number of web pages// first line is the URL of the web site// links to other pages are in the form <A HREF="url">// last line of website is </HTML>// following the websites are a series of url pairs ending with// "The End". // Output links from A to B// Statements if you can surf from 1st to 2nd url in the pairs.  import java.awt.*;import hsa.*; public class J5Surfin{    static Console c;               public static String [] url = new String [100];      public static void main (String [] args)    {	c = new Console ();	boolean [] [] adj = new boolean [100] [100];	String line, to, a, b;	int n, x, y, k; 	TextInputFile fi = new TextInputFile ("surf.in2");	TextOutputFile fo = new TextOutputFile ("surf.ou2"); 	for (int i = 0 ; i < 100 ; i++)	    for (int j = 0 ; j < 100 ; j++)	    {		if (i == j)		    adj [i] [j] = true;		else		    adj [i] [j] = false;	    } 	// get the url's	n = fi.readInt ();	for (int i = 0 ; i < n ; i++)	{	    url [i] = fi.readLine ();	    line = fi.readLine ();	    while (!line.equals ("</HTML>"))		line = fi.readLine ();	}	fi.close (); 	// get the links	fi = new TextInputFile ("surf.in2");	for (int i = 0 ; i < n ; i++)	{	    line = fi.readLine ();	    line = fi.readLine ();	    while (!line.equals ("</HTML>"))	    {		x = line.indexOf ("<A HREF=\"");		while (x >= 0)		{		    y = line.indexOf ("\">", x + 9);		    to = line.substring (x + 9, y);		    fo.println ("Link from " + url [i] + " to " + to);		    c.println ("Link from " + url [i] + " to " + to);		    k = find (to, n);		    adj [i] [k] = true;		    line = line.substring (y + 2);		    x = line.indexOf ("<A HREF=\"");		}		line = fi.readLine ();	    }	} 	// Transitive closure	warshalls (adj); 	// read and print "can or can't surf"s, using the adjacency matrix	a = fi.readLine ();	fo.println ();	c.println ();	while (!a.equals ("The End"))	{	    b = fi.readLine ();	    if (adj [find (a, n)] [find (b, n)])	    {		fo.println ("Can surf from " + a + " to " + b);		c.println ("Can surf from " + a + " to " + b);	    }	    else	    {		fo.println ("Can't surf from " + a + " to " + b);		c.println ("Can't surf from " + a + " to " + b);	    }	    a = fi.readLine ();	}	fi.close ();	fo.close ();    }      // linear sreach for s in the url array.    // this should always be found.    static public int find (String s, int n)    {	int k = 0;	while (k < n && !url [k].equals (s))	    k++;	if (k == n)	    c.println ("Error: can't find " + s);	return k;    }      // Warshall's classic transitive closure algorithm.    static public void warshalls (boolean [] [] a)    {	for (int y = 0 ; y < 100 ; y++)	    for (int x = 0 ; x < 100 ; x++)		if (a [x] [y])		    for (int j = 0 ; j < 100 ; j++)			if (a [y] [j])			    a [x] [j] = true;    }}

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