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Problem solution · Java

CCC 2003 S3 - Floor Plan

CCC 2003 S3 - Floor Plan: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Depth-first search
Source
CCCSolutions
Length
119 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

Depth-first search

For CCC 2003 S3 - Floor Plan, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 119 lines of Java from the credited upstream file ccc03s3.java.
  • The implementation visibly relies on sequence storage.
  • 8 loop blocks detected.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 2003 S3 - Floor Plan · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2003// Problem S3J5 Floor Plan//// given a grid with walls and an amount of flooring,// flooring the larger rooms first, determine the flooring reamining// and the number of rooms floored.//// 1. read the floor plan (rooms are 0, walls are -1)// 2. use recusion to determine which squares belong to a single room//    (number the first room 1, second 2, etc.)// 3. determine the size of each room (add up all the 1's, 2's etc.)// 4. find largest room, and floor it, set its value to -1, repeat! // file input: amount of flooring, row, col and grid given// output the # rooms floored and the remianing flooring import java.awt.*;import hsa.*; public class S3J5Floor{    static Console cc;    static int [] [] house;    static int r, c;     public static void main (String [] args)    {	cc = new Console (); 	String line;	int n, k;	int [] room;	int count, largest;	boolean done; 	TextInputFile fi = new TextInputFile ("floor5.in");	TextOutputFile fo = new TextOutputFile ("floor5.out"); 	// read grid	// put -1 for walls, 0 for room	n = fi.readInt ();	r = fi.readInt ();	c = fi.readInt ();	house = new int [r] [c];	for (int i = 0 ; i < r ; i++)	{	    line = fi.readLine ();	    for (int j = 0 ; j < c ; j++)		if (line.charAt (j) == 'I')		    house [i] [j] = -1;		else		    house [i] [j] = 0;	} 	//number the rooms	k = 1;	for (int i = 0 ; i < r ; i++)	    for (int j = 0 ; j < c ; j++)		if (house [i] [j] == 0)		{		    check (i, j, k);		    k++;		} 	// determine area of rooms	room = new int [500];	for (int i = 0 ; i < r ; i++)	    for (int j = 0 ; j < c ; j++)		if (house [i] [j] > 0)		    room [house [i] [j]]++; 	// get next largest room and floor it	count = 0;	done = false;	while (!done && n > 0)	{	    largest = 0;	    for (int i = 0 ; i < 500 ; i++)		if (room [i] > room [largest])		    largest = i;	    if (room [largest] > 0)	    {		if (room [largest] <= n)		{		    n = n - room [largest];		    room [largest] = -1;		    count++;		}		else		    done = true;	    }	    else		done = true;	} 	fo.println (count + " rooms, " + n + " square metre(s) left over"); 	fi.close ();	fo.close ();    }      // sets the house grid are current location to k, and recursively    // sets all connected square to k as well.    public static void check (int i, int j, int k)    {	if (i >= 0 && i < r && j >= 0 && j < c && house [i] [j] == 0)	{	    house [i] [j] = k;	    check (i - 1, j, k);	    check (i + 1, j, k);	    check (i, j + 1, k);	    check (i, j - 1, k);	}    }}

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