[英]Maze solving using recursion in Java
作业问题在这里,已经花了几个小时和几个小时编码,但是我似乎无法获得正确的输出,因此我希望从你们那里获得一些帮助。
我的代码是:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
public class Maze {
private int[][] grid;
private final int TRIED = 2;
private final int PATH = 3;
public Maze(String fileName) throws IOException {
BufferedReader in = new BufferedReader(new FileReader(fileName));
String s;
String [] aux= new String [2];
int x=0;
aux=in.readLine().split("\\s*[, .]\\s*");
int dimensao=Integer.parseInt(aux[0]);
String [][] gridString= new String [dimensao][dimensao];
grid= new int[dimensao][dimensao];
while ((s=in.readLine())!=null){
gridString [x]=s.split("\\s*[, .]\\s*");
x++;
}
for (int i=0;i<gridString.length;i++){
for(int j=0;j<gridString[0].length;j++){
switch (gridString[i][j]) {
case "true":
grid[i][j]=1;
break;
case "false":
grid[i][j]=0;
break;
}
}
}
}
public boolean traverse(int row, int column) {
boolean done = false;
if (valid(row, column)) {
grid[row][column] = TRIED; // this cell has been tried
if (row==grid.length-1)
done = true; // the maze is solved
else {
done = traverse(row + 1, column); // down
if (!done)
done = traverse(row, column + 1); // right
if (!done)
done = traverse(row - 1, column); // up
if (!done)
done = traverse(row, column - 1); // left
}
if (done) // this location is part of the final path
grid[row][column] = PATH;
}
return done;
}
private boolean valid(int row, int column) {
boolean result = false;
if (row >= 0 && row < grid.length
&& column >= 0 && column < grid[row].length) // check if cell is not blocked and not previously tried
{
if (grid[row][column] == 1) {
result = true;
}
}
return result;
}
public String toString (){
String result = "\n";
for (int row = 0; row < grid.length; row++) {
for (int column = 0; column < grid[0].length; column++) {
result += grid[row][column] + "";
}
result += "\n";
}
return result;
}
public String [][]transformMatrix (int [][]array ){
String [][] matrizChar= new String [array.length][array[0].length];
for (int i=0; i<array.length; i++)
for(int j=0; j<array[0].length;j++){
if (array[i][j]==1)
matrizChar[i][j]="*";
else matrizChar[i][j]="-";
}
return matrizChar;
}
}
另一类是:
public static void main(String[] args) throws IOException {
{
Maze labyrinth = new Maze("rede.txt");
System.out.println(labyrinth);
boolean done=labyrinth.traverse(0, 0);
if (labyrinth.traverse(0, 0)) {
System.out.println("The maze was successfully traversed!");
} else {
System.out.println("There is no possible path.");
}
System.out.println(labyrinth);
}
}
}
我正在使用的文字输入是
五
正确,错误,正确,错误,错误
真实,错误,真实,真实,真实
错误,错误,错误,错误,真实
真实,真实,错误,真实,真实
错误,错误,错误,真实,错误
并生成数组:
1 0 1 0 0
1 0 1 1 1
0 0 0 0 1
0 0 0 1 0
0 0 0 1 0
因为它将输入文本文件转换为整数数组。
现在我的问题是它可以像这样解决迷宫
2 0 1 0 0
2 0 1 1 1
0 0 0 0 1
1 1 0 1 1
0 0 0 1 0
什么时候应该这样解决
2 0 3 0 0
2 0 3 3 3
0 0 0 0 3
1 1 0 3 3
0 0 0 3 0
它还说迷宫没有解决方案,因此我猜问题出在“遍历”方法上,但我真诚地找不到错误所在。 非常感谢您的帮助!
看起来您的问题出在traverse()
方法内部的return语句中。 通过此过程:
该程序
public boolean traverse(int row, int column) {
boolean done = false;
if (valid(row, column)) {
grid[row][column] = TRIED; // this cell has been tried
if (row==grid.length-1)
done = true; // the maze is solved
else {
done = traverse(row + 1, column); // not valid
if (!done)
done = traverse(row, column + 1); // not valid
if (!done)
done = traverse(row - 1, column); // not valid
if (!done)
done = traverse(row, column - 1); // not valid
}
if (done) // this location is part of the final path
grid[row][column] = PATH;
}
return done;
}
还值得注意
您的起始位置(0,0)没有到达终点的合法路径,因此您自然会得到答案。 如果要查找合法路径,则遍历所有可能的起点或将起点设置在实际可完成路径上的某个位置。
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