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CCC 1998 P5 - Mountain Passage

CCC 1998 P5 - Mountain Passage: 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
204 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 1998 P5 - Mountain Passage, 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

  • 204 lines of Java from the credited upstream file ccc98s5.java.
  • The implementation visibly relies on sequence storage.
  • 7 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 1998 P5 - Mountain Passage · JavaJava
Use this to learn the idea, then write your own version.
// CCC 1998// Problem E: MountianPassage // Can't go up or down more than 2 units, if either location > top left// corner, oxygen is reguired. // this is a dynamic programmming type problem:// starting at top corner, create a list of all points visited (the top one)// got thru the list and update a distance array (a 2D array)// (the distance is the same as current or 1 more, provided you can reach it.)// create a new list of all the newly changed points.// continue until the list is empty. // eg: if you had// 2 3 4// 4 6 9// 6 9 5//// the list (x) would be (0,0)// distance array would be:// 0    9999 9999// 9999 9999 9999// 9999 9999 9999 // then the new list (y) would be (0,1) and (1,0)// distance array would be:// 0       1 9999// 1    9999 9999// 9999 9999 9999 // copy y into x// then the new list (y) would be (0,2), (1,1) and (2,0)// distance array would be:// 0    1    2// 1    2 9999// 2 9999 9999 // copy y into x// then the new list (y) would be null// can't go any further // BUT if you could, reading the bottom right element gives the answer. // file input and output.// input:  number of trips, then size of square and the then the square itself// output: the # of oxygen units used. import java.awt.*;import hsa.*; public class P5MountianPassage{    static Console cc;     public static void main (String [] args)    {	cc = new Console ();	TextInputFile fi = new TextInputFile ("passage.in");	TextOutputFile fo = new TextOutputFile ("passage.out");	int n, oxy;	int [] [] g;	int size; 	n = fi.readInt ();	for (int i = 1 ; i <= n ; i++)	{  	    // read the size,	    // create the main grid (g) and distance (d) 2D arrays	    size = fi.readInt ();	    g = new int [size] [size];	    int [] [] d = new int [size] [size];	    for (int r = 0 ; r < size ; r++)		for (int c = 0 ; c < size ; c++)		{		    g [r] [c] = fi.readInt ();		    d [r] [c] = 9999;		} 	    // create the old (x) and new(y) lists of points	    Point [] x = new Point [625];	    Point [] y = new Point [625];	    for (int j = 0 ; j < 625 ; j++)	    {		x [j] = new Point ();		y [j] = new Point ();	    } 	    // set oxygen limit and initialize list (x) and d	    oxy = g [0] [0];	    x [0].r = 0;	    x [0].c = 0;	    d [0] [0] = 0;	    int k = 1;  // size of old list (x)	    int h;	    do	    { 		// t is the size of the new list (y)		int t = 0; 		for (int j = 0 ; j < k ; j++)		{		    int a = x [j].r;		    int b = x [j].c; 		    // go up if you can		    if (a - 1 >= 0 && d [a - 1] [b] > d [a] [b] && Math.abs (g [a - 1] [b] - g [a] [b]) <= 2)		    {			h = d [a - 1] [b];			if (g [a - 1] [b] > oxy || g [a] [b] > oxy)			    d [a - 1] [b] = d [a] [b] + 1;			else			    d [a - 1] [b] = d [a] [b];			if (d [a - 1] [b] < h)			{			    y [t].r = a - 1;			    y [t].c = b;			    t++;			}		    } 		    // go left if you can		    if (b - 1 >= 0 && d [a] [b - 1] > d [a] [b] && Math.abs (g [a] [b - 1] - g [a] [b]) <= 2)		    {			h = d [a] [b - 1];			if (g [a] [b - 1] > oxy || g [a] [b] > oxy)			    d [a] [b - 1] = d [a] [b] + 1;			else			    d [a] [b - 1] = d [a] [b];			if (d [a] [b - 1] < h)			{			    y [t].r = a;			    y [t].c = b - 1;			    t++;			}		    } 		    // go down if you can		    if (a + 1 < size && d [a + 1] [b] > d [a] [b] && Math.abs (g [a + 1] [b] - g [a] [b]) <= 2)		    {			h = d [a + 1] [b];			if (g [a + 1] [b] > oxy || g [a] [b] > oxy)			    d [a + 1] [b] = d [a] [b] + 1;			else			    d [a + 1] [b] = d [a] [b];			if (d [a + 1] [b] < h)			{			    y [t].r = a + 1;			    y [t].c = b;			    t++;			}		    } 		    // go right if you can		    if (b + 1 < size && d [a] [b + 1] > d [a] [b] && Math.abs (g [a] [b + 1] - g [a] [b]) <= 2)		    {			h = d [a] [b + 1];			if (g [a] [b + 1] > oxy || g [a] [b] > oxy)			    d [a] [b + 1] = d [a] [b] + 1;			else			    d [a] [b + 1] = d [a] [b];			if (d [a] [b + 1] < h)			{			    y [t].r = a;			    y [t].c = b + 1;			    t++;			}		    }		} 		// move new (y) into old (x) list		for (int j = 0 ; j < t ; j++)		{		    x [j].r = y [j].r;		    x [j].c = y [j].c;		}		k = t;	    }	    while (k > 0); 	    // the answer is in the bottom right of distance array	    if (d [size - 1] [size - 1] == 9999)	    {		fo.println ("CANNOT MAKE TRIP");		cc.println ("CANNOT MAKE TRIP");	    }	    else	    {		fo.println (d [size - 1] [size - 1]);		cc.println (d [size - 1] [size - 1]);	    }	    fo.println ();	}     }} class Point{    public int r, c;}

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