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2013
11-09

POJ 1522 N-Credible Mazes [解题报告] Java

N-Credible Mazes

问题描述 :

An n-tersection is defined as a location in n-dimensional space, n being a positive integer, having all non-negative integer coordinates. For example, the location (1,2,3) represents an n-tersection in three dimensional space. Two n-tersections are said to be adjacent if they have the same number of dimensions and their coordinates differ by exactly 1 in a single dimension only. For example, (1,2,3) is adjacent to (0,2,3) and (2,2,3) and (1,2,4), but not to (2,3,3) or (3,2,3) or (1,2). An n-teresting space is defined as a collection of paths between adjacent n-tersections.

Finally, an n-credible maze is defined as an n-teresting space combined with two specific n-tersections in that space, one of which is identified as the starting n-tersection and the other as the ending n-tersection.

输入:

The input file will consist of the descriptions of one or more n-credible mazes. The first line of the description will specify n, the dimension of the n-teresting space. (For this problem, n will not exceed 10, and all coordinate values will be less than 10.) The next line will contain 2n non-negative integers, the first n of which describe the starting n-tersection, least dimension first, and the next n of which describe the ending n-tersection. Next will be a nonnegative number of lines containing 2n non-negative integers each, identifying paths between adjacent n-tersections in the n-teresting space. The list is terminated by a line containing only the value �. Several such maze descriptions may be present in the file. The end of the input is signalled by space dimension of zero. No further data will follow this terminating zero.

输出:

For each maze output it’s position in the input; e.g. the first maze is “Maze #1″, the second is “Maze #2″, etc. If it is possible to travel through the n-credible maze’s n-teresting space from the starting n-tersection to the ending n-tersection, also output “can be travelled” on the same line. If such travel is not possible, output “cannot be travelled” instead.

样例输入:

2 
0 0 2 2
0 0 0 1
0 1 0 2
0 2 1 2
1 2 2 2
-1
3
1 1 1 1 2 3
1 1 2 1 1 3
1 1 3 1 2 3
1 1 1 1 1 0
1 1 0 1 0 0
1 0 0 0 0 0
-1
0

样例输出:

Maze #1 can be travelled
Maze #2 cannot be travelled

解题代码:

/* @author:zeropinzuo */
import java.io.*;
import java.util.*;

public class Main{
 static Scanner cin;
 static int count=0;
 public static void main(String args[]){
  cin = new Scanner(System.in);
  int n;
  while((n=cin.nextInt())!=0)
   run(n);
  }
	
 static void run(int n){
   Point start,end;
   start = new Point(n, cin);
   end   = new Point(n, cin);
		
   ArrayList< PointSet> Mset = new ArrayList< PointSet>();
		
   PointSet startSet = new PointSet();
   PointSet endSet = new PointSet();
   startSet.add(start);
   endSet.add(end);
	
   Mset.add(startSet);
   Mset.add(endSet);
		
   Point A,B;
   PointSet setA=new PointSet(), setB=new PointSet();
   while(true){
    if(cin.hasNext("-1") == true){
	 cin.nextInt();
	 break;
    }
				
    setA = null;
    setB = null;
    A = new Point(n, cin);
    B = new Point(n, cin);
			
    for(PointSet set:Mset){
	if(set.contains(A))
	  setA = set;
	if(set.contains(B))
	  setB = set;
    }
			
   if((setA != null)&&(setB != null)){
    if(setA != setB){
	setA.addAll(setB);
	Mset.remove(setB);
    }	
   }
   else if(setA != null)
    setA.add(B);
   else if(setB != null)
    setB.add(A);
   else{
    setA = new PointSet();
    setA.add(A);
    setA.add(B);
    Mset.add(setA);
   }
										
 }
		
 startSet = null;
 endSet   = null;
 for(PointSet set:Mset){
  if(set.contains(start))
    startSet = set;
  if(set.contains(end))
    endSet = set;
  }
		
  if(startSet == endSet)
    System.out.println("Maze #"+(++count)+" can be travelled");
  else
    System.out.println("Maze #"+(++count)+" cannot be travelled");
 }
}

class Point implements Comparable{
 int num;
		
 public Point(int n, Scanner cin){
   num=0;
 for(int i=0;i< n;i++)
  num = 10*num+cin.nextInt();
}
	
 public int compareTo(Object another){
   Point p = (Point)another;
	
   return this.num-p.num;
 }
}

class PointSet extends TreeSet< Point>{
	
}

  1. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
    O(n) = O(n-1) + O(n-2) + ….
    O(n-1) = O(n-2) + O(n-3)+ …
    O(n) – O(n-1) = O(n-1)
    O(n) = 2O(n-1)