广度优先搜索8平铺游戏

时间:2017-04-28 21:46:13

标签: java breadth-first-search

我创建了一个基于广度优先搜索的8平铺游戏。让我解释一下游戏:

您将获得一个包含8个整数的数组:

012
348
675

并给出一个整数m,它告诉你哪个瓷砖是空白的,在这个例子中我们说它将是8.你只能将这个瓷砖与瓷砖直接在其上方,下方,从右到左交换,使用这个您需要对数组进行排序的方法,如下所示:

012
345
678

您需要将空白区域的步骤放入向量中。因此,在上面的例子中,向量将是5,8,因为空白空间从位置#5开始,并在排序时结束于位置#8。我创建了排序代码,现在我无法弄清楚如何将位置添加到向量中,这是我的代码:

package application;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Vector;

public class Solution {

    /******************************************
     * Implementation Here
     ***************************************/

    /*
     * Implementation here: you need to implement the Breadth First Search
     * Method
     */
    /* Please refer the instruction document for this function in details */

    public static LinkedHashSet<int[]> OPEN = new LinkedHashSet<int[]>();
    public static HashSet<int[]> CLOSED = new HashSet<int[]>();
    public static boolean STATE = false;
    public static int empty;

    public static void breadthFirstSearch(int[] num, int m, Vector solution1) {

        //int[] start = num;
        int[] start = {0,1,2,3,4,5,6,8,7};
        m = 7;
        int[] goal = { 0, 1, 2, 3, 4, 5, 6, 7, 8 };
        int[] X;
        int[] temp = {};

        OPEN.add(start);

        while (OPEN.isEmpty() == false) {


            X = OPEN.iterator().next();
            OPEN.remove(X);

            for (int i = 0; i < X.length; i++) {
                if (X[i] == m) {
                    empty = i;
                    System.out.println("empty = " + empty);
                    solution1.addElement(empty);
                }
            }

            // get position of ZERO or EMPTY SPACE
            if (compareArray(X, goal)) {


                System.out.println("SUCCESS");
                for(Object i : solution1) {
                    System.out.println((int)i);
                }
                print(X);
                break;
            } else {


                //print(X);
                System.out.println("------");

                // generate child nodes
                CLOSED.add(X);
                int[][] y = new int[4][9];
                for (int i = 0; i < 4; i++)
                    for (int i1 = 0; i1 < 9; i1++)
                        y[i][i1] = X[i1];

                temp = up(y[0], empty);
                if (temp != null) {
                    OPEN.add(temp);
                    System.out.println("doing up");

                    System.out.println("added "+ empty);
                //  print(temp);

                }

                temp = left(y[1], empty);

                if (temp != null) {
                    OPEN.add(temp);
                    System.out.println("doing left");

                    System.out.println("added "+ empty);
            //      print(temp);
                }

                temp = down(y[2], empty);
                if (temp != null) {
                    OPEN.add(temp);
                    System.out.println("doing down");

                    System.out.println("added "+ empty);
            //      print(temp);
                }
                temp = right(y[3], empty);
                if (temp != null) {
                    OPEN.add(temp);
                    System.out.println("doing right");

                    System.out.println("added "+ empty);
            //      print(temp);

                }
                if (OPEN.isEmpty()) {
                    // System.out.println("Ending loop");
                }
            }
        }

    }

    public static void print(int[] arr) {
        System.out.println(arr[0] + " " + arr[1] + " " + arr[2]);
        System.out.println(arr[3] + " " + arr[4] + " " + arr[5]);
        System.out.println(arr[6] + " " + arr[7] + " " + arr[8]);
        System.out.println("---------");

    }

    public static boolean compareArray(int[] a, int[] b) {

        for (int i = 0; i < a.length; i++)
            if (a[i] != i)
                return false;

        return true;

    }

    public static int[] up(int[] s, int p) {
        boolean b = false;
        int[] str = s;
        if (p > 3) {
            int temp = str[p - 3];
            str[p - 3] = str[p];
            str[p] = temp;
            b = true;

        }
        // Eliminates child of X if its on OPEN or CLOSED
        int[] t = new int[9];

        for (int i = 0; i < str.length; i++)
            t[i] = str[i];

        if (!OPEN.contains(t) && CLOSED.contains(t) == false && b)
            return str;
        else {
            System.out.println("returning null for up");
            return null;
        }
    }

    public static int[] down(int[] s, int p) {

        int[] str = s;
        boolean b = false;
        if (p < 6) {
            int temp = str[p + 3];
            str[p + 3] = str[p];
            str[p] = temp;

            b = true;
        }
        int[] t = new int[9];
        for (int i = 0; i < str.length; i++)
            t[i] = str[i];

        if (!OPEN.contains(str) && CLOSED.contains(str) == false && (b))
            return str;
        else {
            System.out.println("returning null for down");
            return null;
        }
    }

    public static int[] left(int[] s, int p) {
        int[] str = s;
        boolean b = false;
        if (p != 0 && p != 3 && p != 6) {
            int temp = str[p - 1];
            str[p - 1] = str[p];
            str[p] = temp;
            b = true;
        }
        // Eliminates child of X if its on OPEN or CLOSED
        int[] t = new int[9];
        for (int i = 0; i < str.length; i++)
            t[i] = str[i];

        if (!OPEN.contains(str) && CLOSED.contains(str) == false && (b))
            return str;
        else {
            System.out.println("returning null for left");
            return null;
        }
    }

    public static int[] right(int[] s, int p) {
        boolean b = false;
        int[] str = s;
        if (p != 2 && p != 5 && p != 8) {
            int temp = str[p + 1];
            str[p + 1] = str[p];
            str[p] = temp;
            b = true;
        }
        // Eliminates child of X if its on OPEN or CLOSED
        int[] t = new int[9];
        for (int i = 0; i < str.length; i++)
            t[i] = str[i];

        if (!OPEN.contains(t) && CLOSED.contains(t) == false && (b))
            return str;
        else {
            System.out.println("returning null for right");
            return null;
        }
    }

}

1 个答案:

答案 0 :(得分:0)

此代码与我最近回答的代码完全相同,请检查this并相应地更新代码。