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CCC 2010 S4 - Animal Farm

CCC 2010 S4 - Animal Farm: 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
173 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 2010 S4 - Animal Farm, 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

  • 173 lines of Java from the credited upstream file ccc10s4.java.
  • The implementation visibly relies on sequence storage.
  • 15 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 2010 S4 - Animal Farm · JavaJava
Use this to learn the idea, then write your own version.
// S4 2010 Animal Farm//// 2D arrays, graphs and Prim's algorithm// to find the Minimal Spanning Tree (MST)//// this is a huge input processing problem// The input has to be converted to a proper// weighted graph, with the vertices being the// ANIMALS themselves. (The last vertix is// the OUTSIDE.)//// once this is done you need calculate the// MST for the graph, TWICE. Once without considering// the outside, and one considering the outside.//// the version of Prim's I'm using came form// "Algorithms" by Robert Sedgewick, 2nd Edition// import java.awt.*;import hsa.*; public class S42010{     public static int[] [] graph;      public static void main (String[] args)    {	TextInputFile c;	c = new TextInputFile ("s4.1.in");	int n;	Info[] [] input; 	int numEdges;	int[] corner = new int [1001];	int[] cost = new int [1001]; 	n = c.readInt (); 	// create and initialize graph	graph = new int [n + 1] [n + 1];	for (int i = 0 ; i < n + 1 ; i++)	    for (int j = 0 ; j < n + 1 ; j++)		graph [i] [j] = 999999999;	for (int j = 0 ; j < n + 1 ; j++)	    graph [j] [j] = 0; 	// initialize the input grid	input = new Info [1001] [1001];	for (int i = 0 ; i < 1001 ; i++)	    for (int j = 0 ; j < 1001 ; j++)		input [i] [j] = new Info (-1, 0); 	// input section	for (int i = 0 ; i < n ; i++)	{	    numEdges = c.readInt ();	    for (int k = 0 ; k < numEdges ; k++)		corner [k] = c.readInt ();	    for (int k = 0 ; k < numEdges ; k++)		cost [k] = c.readInt (); 	    for (int k = 0 ; k < numEdges ; k++)	    {		int j = (k + 1) % numEdges; 		// and old fence		if (input [corner [k]] [corner [j]].cost > 0)		{ 		    // this if is needed for test data 3,4 and 5. In these cases		    // there are TWO or more fences between the same two pens with		    // different costs. Take only the smaller one.		    if (graph [i] [input [corner [k]] [corner [j]].animal] > cost [k])		    {			graph [i] [input [corner [k]] [corner [j]].animal] = cost [k];			graph [input [corner [k]] [corner [j]].animal] [i] = cost [k];		    }		    input [corner [k]] [corner [j]].animal = -1;		    input [corner [j]] [corner [k]].animal = -1;		} 		// a new fence		else		{		    input [corner [k]] [corner [j]].cost = cost [k];		    input [corner [k]] [corner [j]].animal = i;		    input [corner [j]] [corner [k]].cost = cost [k];		    input [corner [j]] [corner [k]].animal = i;		}	    }	} 	// at this point, only outside edges in input array	// have >0 in the animal field.	// Need to get the cost for each animal to outside	// (that is, to animal[n])	for (int i = 0 ; i < 1001 ; i++)	    for (int j = 0 ; j < 1001 ; j++)		if (input [i] [j].animal >= 0)		{		    if (graph [input [i] [j].animal] [n] > input [i] [j].cost)		    {			graph [input [i] [j].animal] [n] = input [i] [j].cost;			graph [n] [input [i] [j].animal] = input [i] [j].cost;		    }		} 	// do Prim's for all animals (excluding outside)	int answer1 = prims (n); 	// do Prim's for all animals (including outside)	int answer2 = prims (n + 1); 	System.out.println (Math.min (answer1, answer2));    }      public static int prims (int n)    {	int unseen = 99999999;	int[] val = new int [n + 1];	for (int k = 1 ; k <= n ; k++)	    val [k] = -unseen;	val [0] = -(unseen + 1);	int min = 1;	int k;	do	{	    k = min;	    val [k] = -val [k];	    min = 0;	    if (val [k] == unseen)		val [k] = 0;	    for (int t = 1 ; t <= n ; t++)	    {		if (val [t] < 0)		{		    if (graph [k - 1] [t - 1] != 999999999 && val [t] < -(graph [k - 1] [t - 1]))			val [t] = -(graph [k - 1] [t - 1]);		    if (val [t] > val [min])			min = t;		}	    }	}	while (min != 0); 	// val contains all the values in the MST	// add them up for the total value of trampled fences	int answer = 0;	for (int i = 1 ; i < n + 1 ; i++)	    answer = answer + val [i];	return answer;    }}  class Info{    public int animal;    public int cost;     public Info (int a, int c)    {	animal = a;	cost = c;    }}

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