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Problem solution · Java

CCC 1999 P4 - A Knightly Pursuit

CCC 1999 P4 - A Knightly Pursuit: 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
154 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 1999 P4 - A Knightly Pursuit, 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

  • 154 lines of Java from the credited upstream file ccc99s4.java.
  • The implementation visibly relies on sequence storage.
  • 9 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 1999 P4 - A Knightly Pursuit · JavaJava
Use this to learn the idea, then write your own version.
// CCC 1999// Problem 4: A Knightly Pursuit // This is a Dynamic programming approach. Given the starting location// of the knight each square will be given a number indicating the minimum// moves needed to get there. (if a square can be reached in k moves// it can also be reached in k + 2x moves, x = 1,2,3...) // -1 indicates the knight can't get there. // file input and output.// input:  number of games, r,c,pr,pc,kr,kr (size of board, pawn/knight pos) import java.awt.*;import hsa.*; public class P4KnightDP{    static Console cc;    static int maxr, maxc;    static int m, ps, nps;    static int [] [] b; // board holds min number of moves to get here    static Point [] p, np; // list of squares to use for next pass (& new list)     public static void main (String [] args)    {	cc = new Console ();	TextInputFile fi = new TextInputFile ("knight.in");	TextOutputFile fo = new TextOutputFile ("knight.out");	int n, hpr, pc, kr, kc;	boolean win, stalemate; 	n = fi.readInt ();	for (int i = 1 ; i <= n ; i++)	{	    maxr = fi.readInt ();	    maxc = fi.readInt ();	    hpr = fi.readInt () - 1;	    pc = fi.readInt () - 1;	    kr = fi.readInt () - 1;	    kc = fi.readInt () - 1; 	    // initialize board and p array	    b = new int [maxr] [maxc];	    for (int r = 0 ; r < maxr ; r++)		for (int c = 0 ; c < maxc ; c++)		    b [r] [c] = -1;	    b [kr] [kc] = 0; 	    p = new Point [maxr * maxc];	    np = new Point [maxr * maxc];	    for (int r = 0 ; r < maxr * maxc ; r++)	    {		p [r] = new Point ();		np [r] = new Point ();	    }	    ps = 1;	    p [0].r = kr;	    p [0].c = kc; 	    // fill board (while p has points to consider.)	    while (ps > 0)	    {		nps = 0;		for (int j = 0 ; j < ps ; j++)		{		    newPoint (p [j].r + 1, p [j].c + 2, p [j].r, p [j].c);		    newPoint (p [j].r - 1, p [j].c + 2, p [j].r, p [j].c);		    newPoint (p [j].r - 2, p [j].c + 1, p [j].r, p [j].c);		    newPoint (p [j].r - 2, p [j].c - 1, p [j].r, p [j].c);		    newPoint (p [j].r - 1, p [j].c - 2, p [j].r, p [j].c);		    newPoint (p [j].r + 1, p [j].c - 2, p [j].r, p [j].c);		    newPoint (p [j].r + 2, p [j].c - 1, p [j].r, p [j].c);		    newPoint (p [j].r + 2, p [j].c + 1, p [j].r, p [j].c);		}		for (int j = 0 ; j < nps ; j++)		{		    p [j].r = np [j].r;		    p [j].c = np [j].c;		}		ps = nps;	    } 	    // check for win	    // win if knight can get to the same spot as pawn	    // eg. a win is if the board shows 3 and the pawn	    //     gets there in 3, 5 or 7 moves.	    // m is the move number. pawn moves first	    m = 1;	    win = false;	    // go < maxr-1 because if p gets to maxr-1 knight loses	    for (int pr = hpr + 1 ; pr < maxr - 1 && !win ; pr++)	    {		if (m >= b [pr] [pc] && b [pr] [pc] >= 0 && (m - b [pr] [pc]) % 2 == 0)		{		    win = true;		    fo.println ("Win in " + m + " knight moves(s).");		    cc.println ("Win in " + m + " knight moves(s).");		}		m++;	    } 	    if (!win)	    {		// check for stalemate		// this is where the knight can get to the square 		// above the pawn. 		// eg. board shows 3 above and pawn gets to the		// square below in 3, 5 or 7 moves		// if knight starts above pawn, its stalemate		// in 0 moves, hence m starts at 0.		m = 0;		stalemate = false;		for (int pr = hpr ; pr < maxr - 1 && !stalemate ; pr++)		{		    if (m >= b [pr + 1] [pc] && b [pr + 1] [pc] >= 0 && (m - b [pr + 1] [pc]) % 2 == 0)		    {			stalemate = true;			fo.println ("Stalemate in " + m + " knight moves(s).");			cc.println ("Stalemate in " + m + " knight moves(s).");		    }		    m++;		}		if (!stalemate)		{		    fo.println ("Loss in " + (maxr - hpr - 2) + " knight moves(s).");		    cc.println ("Loss in " + (maxr - hpr - 2) + " knight moves(s).");		}	    }	}	fo.close ();	fi.close ();    }      public static void newPoint (int r, int c, int or, int oc)    {	if (r >= 0 && r < maxr && c >= 0 && c < maxc && b [r] [c] == -1)	{	    b [r] [c] = b [or] [oc] + 1;	    np [nps].r = r;	    np [nps].c = c;	    nps++;	}    }} class Point{    public int r, c;}

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