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Problem solution · Java

CCC 2004 S5 - Super Plumber

CCC 2004 S5 - Super Plumber: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Sliding window or two pointers
Source
CCCSolutions
Length
297 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

Sliding window or two pointers

For CCC 2004 S5 - Super Plumber, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 297 lines of Java from the credited upstream file ccc04s5.java.
  • The implementation visibly relies on sequence storage.
  • 13 loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 2004 S5 - Super Plumber · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2004 // Problem S5 Super Plumber // // A dynamic programming problem: given a 2D array with // coins and obstacles, find max money you can get from bottom // left to bottom right, going only up, down and right. // // Idea is to find the max amount of money to every square in the grid // working left to right, one column at a time, // fill up the max money array.  // at any row, go right (if you can) and up (all the way) // updating money. Then go right and down (all the way) // update money.  // // the answer is read from the array at (rows-1, cols-1).     import java.awt.*; import java.util.*; import hsa.*;   public class S5SuperPlumberDP {     static Console cc;     static int[] [] grid;     static int rows, cols;       public static void main (String[] args)     { 	cc = new Console (); 	String file;   	cc.print ("file name: "); 	file = cc.readString (); 	TextInputFile fi = new TextInputFile (file);   	rows = fi.readInt (); 	cols = fi.readInt (); 	while (rows > 0) 	{ 	    getGrid (fi); 	    cc.println (solve ()); 	    rows = fi.readInt (); 	    cols = fi.readInt (); 	}     }         public static void getGrid (TextInputFile fi)     { 	String line; 	grid = new int [rows] [cols]; 	for (int r = 0 ; r < rows ; r++) 	{ 	    line = fi.readLine (); 	    for (int c = 0 ; c < cols ; c++) 		if (line.charAt (c) == '.') 		    grid [r] [c] = 0; 		else if (line.charAt (c) == '*') 		    grid [r] [c] = -1; 		else 		    grid [r] [c] = line.charAt (c) - '0'; 	}     }         public static int solve ()     { 	int[] [] money; 	money = new int [rows] [cols]; 	for (int i = 0 ; i < rows ; i++) 	    for (int j = 0 ; j < cols ; j++) 		money [i] [j] = -1;   	// establish first column (go up and add them all) 	// stop if there's a star (above is unreachable) 	money [rows - 1] [0] = grid [rows - 1] [0]; 	for (int i = rows - 2 ; i >= 0 ; i--) 	{ 	    if (grid [i] [0] >= 0) 		money [i] [0] = money [i + 1] [0] + grid [i] [0]; 	    else 		i = -1; 	}   	// For each column, go right from each square and 	// work up and down filling in money. 	for (int c = 1 ; c < cols ; c++) 	{ 	    // going down from each square (0 to rows-1) 	    for (int r = 0 ; r < rows ; r++) 	    { 		if (money [r] [c - 1] >= 0) // can't go right from a wall 		{ 		    int t = money [r] [c - 1]; 		    for (int k = r ; k < rows ; k++) 		    { 			if (grid [k] [c] >= 0) 			{ 			    t = t + grid [k] [c]; 			    if (t > money [k] [c]) 				money [k] [c] = t; 			} 			else 			    k = rows; // get out: hit a wall 		    } 		} 	    } 	    // going up from each square (rows-1 to 0) 	    for (int r = rows - 1 ; r >= 0 ; r--) 	    { 		if (money [r] [c - 1] >= 0) // can't go right from a wall 		{ 		    int t = money [r] [c - 1]; 		    for (int k = r ; k >= 0 ; k--) 		    { 			if (grid [k] [c] >= 0) 			{ 			    t = t + grid [k] [c]; 			    if (t > money [k] [c]) 				money [k] [c] = t; 			} 			else 			    k = -1; // get out: hit a wall 		    } 		} 	    }    /*           for (int i = 0 ; i < rows ; i++) 	    { 		for (int j = 0 ; j < cols ; j++) 		{ 		    if (money [i] [j] < 10) 			cc.print (money [i] [j] + "  "); 		    else if (money [i] [j] < 100) 			cc.print (money [i] [j] + " "); 		    else 			cc.print (money [i] [j] + ""); 		} 		cc.println (); 	    } 	    cc.println (); 	    cc.println (); 	    cc.getChar ();  */       } 	return money [rows - 1] [cols - 1];     } }

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