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Problem solution · Java

CCC 2003 S5 - Trucking

CCC 2003 S5 - Trucking: 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
124 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 2003 S5 - Trucking, 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

  • 124 lines of Java from the credited upstream file ccc03s5.java.
  • The implementation visibly relies on sequence storage.
  • 6 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 2003 S5 - Trucking · JavaJava
Use this to learn the idea, then write your own version.
// CCC 2003// Problem S5 Trucking//// given a series of cities and "weights" between them// find minimum weight needed to get to a subset of the cities// starting from city #1. // weights are stored in an adjacency matrix// note if there are two weights between same two cities,// use only the higher one. // starting at 1, create the minimum spanning tree for the// weighted graph, using prim's algorithm// the maximum weight possible to get to each city is then// stored in the val array. Find the smallest of these weights// for the destination city: that's the answer! // file input: number of cities, # of weights and # of destination cites given// then the weights and finally the destination cites are given // thanks to Vassili Skarine for pointing out this method to me :-) import java.awt.*;import hsa.*; public class S5TruckingPrims{    static Console cc;    static int [] [] weights;    static int [] dest;     static int n, r, d, a, b, c, tt;     static int [] val;    static boolean [] visited;    static int max, maxt, k, smallest;      public static void main (String [] args)    {	cc = new Console ();  	TextInputFile fi = new TextInputFile ("truck5.in");	TextOutputFile fo = new TextOutputFile ("truck5a.out"); 	n = fi.readInt ();	r = fi.readInt ();	d = fi.readInt (); 	weights = new int [n + 1] [n + 1];	dest = new int [d]; 	// read the roads and store the largest weight, if duplicates	// (road may come in reversed, for arrange smallest first.)	for (int i = 0 ; i < r ; i++)	{	    a = fi.readInt ();	    b = fi.readInt ();	    c = fi.readInt ();	    if (a > b)	    {		tt = a;		a = b;		b = tt;	    }	    if (c > weights [a] [b])	    {		weights [a] [b] = c;		weights [b] [a] = c;	    }	} 	// read the destination cities that need to be visited	for (int i = 0 ; i < d ; i++)	    dest [i] = fi.readInt (); 	// prim's MST algorithm	val = new int [n + 1];	visited = new boolean [n + 1];	for (k = 0 ; k < n + 1 ; k++)	{	    val [k] = 0;	    visited [k] = false;	} 	val [1] = 100000;	maxt = 1; 	do	{	    k = maxt;	    visited [maxt] = true;	    max = 0;	    maxt = -1;	    for (int t = 1 ; t < n + 1 ; t++)	    {		if (val [t] < Math.min (val [k], weights [k] [t]))		    val [t] = Math.min (val [k], weights [k] [t]);		if (val [t] >= max && !visited [t])		{		    max = val [t];		    maxt = t;		}	    }	}	while (maxt != -1); 	// the answer is the smallest value for the destination cities.	smallest = 100000;	for (int i = 0 ; i < d ; i++)	    if (val [dest [i]] < smallest)		smallest = val [dest [i]]; 	fo.println (smallest);	cc.println (smallest); 	fi.close ();	fo.close ();    }}

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