这里的家庭作业问题,已经编写了几个小时,几小时和几个小时的编码,但我似乎无法得到正确的输出,所以我希望得到你们的一些帮助。
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
5
true,false,true,false,false
true,false,true,true,true
false,false,false,false,true
true,true,false,true,true
false,false,false,true,false
它会生成数组:
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。
现在我的问题是它解决了这样的迷宫
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
什么时候应该解决这个问题
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,所以我猜这个问题依赖于“遍历”方法,但我真的无法找出错误的位置。 非常感谢帮助!
答案 0 :(得分:0)
看起来您的问题来自traverse()
方法中的return语句。走过这个:
该计划
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;
}
同样值得注意
你的起始位置(0,0)没有通往终点的合法路径,所以很自然你会得到它作为答案。如果你想找到合法的路径,那么要么遍历所有可能的起点,要么在实际的可完成路径上的某个地方设置起点。