[英]Boggle Game Board Search Program Issues
我正在編程一個笨拙的游戲板解算器。 它使用堆棧,從.dat / .txt文件讀取的用於游戲板的2D字母陣列數組,並搜索“狀態”以存儲其具體位置(到目前為止的點坐標,單詞)。 它旨在在板上搜索每個可能的字母組合以形成長度為3或更大的字符串,然后檢查字典文件以查看該單詞是否為有效的臨時單詞解決方案。 之后,它存儲單詞並返回參數中給出的游戲板解決方案列表。
我的問題:由於某種原因,該程序使我無法象以前從未寫過的一樣。 我對“狀態”的概念非常陌生,因此這可能是一個潛在的問題。 我相信我所擁有的已經相當接近工作了,我為此感到茫然。 當前的問題是它在檢查相鄰字母時從未存儲當前字母和構建字符串。 它確實會正確檢查鄰居,但不會構建任何字符串。 這是代碼:
BoggleSearch:包含main方法,充當驅動程序類。
import java.io.*;
import java.util.*;
public class BoggleSearch {
protected static int GRID_SIZE = 4;
public static String[][] grid = new String[GRID_SIZE][GRID_SIZE];
public static void main(String[] args) throws FileNotFoundException {
if (args.length != 1) {
System.err.println("Usage: java BoggleSearch gridFile");
System.exit(1);
}
Scanner scan = new Scanner(new File(args[0]));
String bigString = scan.next();
bigString = bigString+scan.next();
bigString = bigString+scan.next();
bigString = bigString+scan.next();
scan.close();
int count = 0;
for (int i = 0; i < GRID_SIZE; i++) {
for (int j = 0; j < GRID_SIZE; j++) {
grid[i][j] = bigString.substring(count, count);
count++;
}
}
WordSearch ws = new WordSearch(grid);
ArrayList<BoggleSearchState> foundWords = ws.startSearch();
System.out.println(foundWords);
}
}
WordSearch:包含所有算法,這些算法可在給定游戲板上找到該字符串的所有可能字母組合,並與字典類進行交叉檢查。
import java.awt.Point;
import java.util.*;
public class WordSearch {
public static Stack<BoggleSearchState> stack;
public static ArrayList<BoggleSearchState> foundWords;
private String[][] grid;
private static final int GRID_SIZE = 4;
public BoggleDictionary dictionary;
public WordSearch(String[][] inputGrid) {
grid = new String[GRID_SIZE][GRID_SIZE];
stack = new Stack<BoggleSearchState>();
foundWords = new ArrayList<BoggleSearchState>();
inputGrid = new String[GRID_SIZE][GRID_SIZE];
try {
dictionary = new BoggleDictionary();
} catch (Exception e) {
System.err.println("blew up while making dict object");
e.printStackTrace();
}
}
public ArrayList<BoggleSearchState> startSearch() {
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid.length; j++) {
BoggleSearchState b = new BoggleSearchState(
new ArrayList<Point>(), grid[i][j]);
Point p = new Point(i, j);
b.path.add(p);
stack.push(b);
while (!stack.isEmpty()) {
BoggleSearchState s = stack.pop();
if (s.getWord().length() >=1 && dictionary.contains(s.getWord())) {
foundWords.add(s);
}
Point loc = s.path.get(s.path.size() - 1);
p = new Point(loc.x,loc.y);
// Bottom Neighbor
if (loc.x + 1 >= 0 && loc.x + 1 < grid.length && loc.y >= 0
&& loc.y < grid.length) {
if (s.getVisited(new Point(p.x+1,p.y)) != true) {
BoggleSearchState neo = new BoggleSearchState(new ArrayList<Point>(),s.getWord() + grid[loc.x + 1][loc.y]);
neo.path.add(new Point(p.x+1,p.y));
stack.push(neo);
}
}
// Top Neighbor
if (loc.x - 1 >= 0 && loc.x - 1 < grid.length && loc.y >= 0
&& loc.y < grid.length) {
if (s.getVisited(new Point(p.x-1,p.y)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x - 1][loc.y]);
neo.path.add(new Point(p.x-1,p.y));
stack.push(neo);
}
}
// Right Neighbor
if (loc.x >= 0 && loc.x < grid.length && loc.y + 1 >= 0
&& loc.y + 1 < grid.length) {
if (s.getVisited(new Point(p.x,p.y+1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x][loc.y + 1]);
neo.path.add(new Point(p.x,p.y+1));
stack.push(neo);
}
}
// Left Neighbor
if (loc.x >= 0 && loc.x < grid.length && loc.y - 1 >= 0
&& loc.y - 1 < grid.length) {
if (s.getVisited(new Point(p.x,p.y-1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x][loc.y - 1]);
neo.path.add(new Point(p.x,p.y-1));
stack.push(neo);
}
}
// Bottom-Right Neighbor
if (loc.x + 1 >= 0 && loc.x + 1 < grid.length
&& loc.y + 1 >= 0 && loc.y + 1 < grid.length) {
if (s.getVisited(new Point(p.x+1,p.y+1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x + 1][loc.y + 1]);
neo.path.add(new Point(p.x+1,p.y+1));
stack.push(neo);
}
}
// Bottom-Left Neighbor
if (loc.x + 1 >= 0 && loc.x + 1 < grid.length
&& loc.y - 1 >= 0 && loc.y - 1 < grid.length) {
if (s.getVisited(new Point(p.x+1,p.y-1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x + 1][loc.y - 1]);
neo.path.add(new Point(p.x+1,p.y-1));
stack.push(neo);
}
}
// Top-Right Neighbor
if (loc.x - 1 >= 0 && loc.x - 1 < grid.length
&& loc.y + 1 >= 0 && loc.y + 1 < grid.length) {
if (s.getVisited(new Point(p.x-1,p.y+1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x - 1][loc.y + 1]);
neo.path.add(new Point(p.x-1,p.y+1));
stack.push(neo);
}
}
// Top-Left Neighbor
if (loc.x - 1 >= 0 && loc.x - 1 < grid.length
&& loc.y - 1 >= 0 && -1 < grid.length) {
if (s.getVisited(new Point(p.x-1,p.y-1)) != true) {
BoggleSearchState neo = new BoggleSearchState(
new ArrayList<Point>(),s.getWord() +
grid[loc.x - 1][loc.y - 1]);
neo.path.add(new Point(p.x-1,p.y-1));
stack.push(neo);
}
}
}
}
}
return foundWords;
}
}
BoggleSearchState:創建一個狀態對象,用於存儲游戲板上字符串形成路徑的每個實例的必要數據。 包含為此目的所必需的方法。
import java.awt.Point;
import java.util.ArrayList;
public class BoggleSearchState {
private String word="";
public ArrayList<Point> path = new ArrayList<Point>();
public BoggleSearchState(ArrayList<Point>path, String word) {
this.path = path;
this.word = word;
}
public String getWord() {
return word;
}
public ArrayList<Point> getLocation() {
return path;
}
public boolean getVisited (Point p) {
ArrayList<Point> newPath = new ArrayList<Point>();
for (Point s: path) {
newPath.add(s);
if (p.equals(s)) {
return true;
}
}
return false;
}
public String toString() {
return this.word;
}
}
BoggleDictionary:為作業指定的編寫得很糟糕的字典類。 它已經過測試並且功能齊全。
// BoggleDictionary.java
import java.io.File;
import java.io.FileInputStream;
import java.io.ObjectInputStream;
import java.io.IOException;
import java.util.Scanner;
import java.util.HashSet;
import java.util.Iterator;
/**
A class that stores a dictionary containing words that can be used in a
Boggle game.
@author Teresa Cole
@version CS221 Fall 2013
*/
public class BoggleDictionary
{
private HashSet<String> dictionary;
/** Create the BoggleDictionary from the file dictionary.dat
*/
@SuppressWarnings("unchecked")
public BoggleDictionary() throws Exception {
ObjectInputStream dictFile = new ObjectInputStream(
new FileInputStream( new File( "dictionary.dat")));
dictionary = (HashSet<String>)dictFile.readObject();
dictFile.close();
}
/** Check to see if a string is in the dictionary to determine whether it
* is a valid word.
* @param word the string to check for
* @return true if word is in the dictionary, false otherwise.
*/
public boolean contains( String word)
{
return dictionary.contains( word);
}
/** Get an iterator that returns all the words in the dictionary, one at a
* time.
* @return an iterator that can be used to get all the words in the
* dictionary.
*/
public Iterator<String> iterator()
{
return dictionary.iterator();
}
/**
Main entry point
*/
static public void main(String[] args)
{
System.out.println( "BoggleDictionary Program ");
Scanner kbd = new Scanner( System.in);
BoggleDictionary theDictionary=null;
try
{
theDictionary = new BoggleDictionary();
}
catch (Exception ioe)
{
System.err.println( "error reading dictionary");
System.exit(1);
}
String word;
/*
while (kbd.hasNext())
{
word = kbd.next();
if (theDictionary.contains( word))
System.out.println( word + " is in the dictionary");
else
System.out.println( word + " is not in the dictionary");
}
*/
Iterator<String> iter = theDictionary.iterator();
while (iter.hasNext())
System.out.println( iter.next());
}
}
我很樂意為此付出任何努力,因為我目前正為此努力。 我知道,有許多方法可以實現其他數據結構或組織方法,以在更有效的運行時中完成此任務。 但是,分配的關注點不是針對效率,而是使用這些數據結構(堆棧等)並了解如何能夠走錯方向,然后安全地回溯並在沒有程序的情況下走新方向的基本原則崩潰。 在此先感謝您,我會盡力回答所有問題。
據我所知,您在WordSearch
有兩個網格,但是將它們都設置為初始化的數組。 您永遠不會使用看起來像內置在main方法中的網格。
但是很難說。
您給了我們很多數據,但信息卻很少。 我們不需要有關程序整體的詳細信息,甚至不需要這個大小的程序。 我們需要知道您的具體問題是什么。 沒有人願意為您調試此程序,實際上很少有人會像我所讀的一樣多。
解決初始化問題后,可調試程序,但可以找出它在何處執行(1)不應執行的操作,以及(2)您不了解的操作。 您需要花費足夠的時間來嘗試自己的#1,以便進行調試方面的學習,這很可能使您可以解釋自己不了解的內容,並能很好地回答特定問題。 “這不是構建字符串”還遠遠不夠; 它不是在構建,在構建字符串是什么意思,等等。我期望這是因為它沒有輸入,但是我沒有做那么多的分析。
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