java - 在java中使用递归来解决迷宫问题

Question

我尝试了这个问题，但由于某种原因，结果不正确。给定一组字符串，找出迷宫中存在多少种可能的解决方案，其中字符串由一个“R”(老鼠)、一个“C”(奶酪)、多个“X”(无法通过的方 block )和“.”(可能的路径)组成。任务是找出老鼠到达奶酪时可以采取的可能路线数，而不会增加其路径上任何点与奶酪之间的(欧几里得)距离。我的代码看起来有什么问题吗？

public class RatRoute { 
private static String[] enc;
private static int count;
private static int[] r;
private static int[] c;

// Test the program
public static void main(String[] args) {
    String[] test = {
            ".R...",
            "..X..",
            "....X",
            "X.X.X",
            "...C."};
    int num1 = numRoutes(test);
    System.out.println(num1);   
}

// Set variables, and call recursive function
public static int numRoutes(String[] enc) {
    RatRoute.enc = enc;     
    r = findR(enc);
    c = findC(enc);
    recursiveHelper(r[0], r[1]);
    return count;
}

// Recursive 
public static void recursiveHelper(int x, int y) {

    /*System.out.println();
    System.out.println();
    for (int k = 0; k < enc.length; k++) {
        System.out.println(enc[k]);
    }*/

    if(isBlock(x,y)) {
        return;
    } else if (isBigger(x,y)) {
        return;
    } else if (isCheese(x, y)) {
        count++;
        //System.out.println("Found the Cheese! Path number: " + count);
        //System.out.println();
        return;
    }

    enc[x] = currentPath(x,y);      
    recursiveHelper(x + 1, y);
    recursiveHelper(x, y + 1);
    recursiveHelper(x, y - 1);
    recursiveHelper(x - 1, y);
    enc[x] = returnPath(x,y);

}

// Change the most recently traveled coordinates into a block 
public static String currentPath(int x, int y) {
    char[] Chars = enc[x].toCharArray();
    Chars[y] = 'X';
    String newString = String.valueOf(Chars);
    return newString;       
}

// Turn path already traveled from blocks back into a usable path to travel (undo the currentPath method)
public static String returnPath(int x, int y) {
    char[] Chars = enc[x].toCharArray();
    Chars[y] = '.';
    String newString = String.valueOf(Chars);
    return newString;       
}

// Check if the next movement is into the cheese
public static boolean isCheese(int x, int y) {
    if (enc[x].charAt(y) == 'C') {
        return true;
    } else {
        return false;
    }
}

// Check if the next movement is into a block, or outside the given array
public static boolean isBlock(int x, int y) {   
    if (x == -1 || y == -1
            || x >= enc.length || y >= enc[x].length()) {
        return true;
    } else if (enc[x].charAt(y) == 'X') {
        //System.out.println(x + "," + y);
        return true;
    } else {
        return false;
    }
}

// See if the distance between the rat and the cheese has gotten larger or smaller
public static boolean isBigger(int x, int y) {
    double rx = r[0]; double ry = r[1];
    double cx = c[0]; double cy = c[1];

    double originalDist = Math.sqrt(Math.pow(rx-cx, 2) + Math.pow(ry-cy, 2));
    double newDist = Math.sqrt(Math.pow(x-cx, 2) + Math.pow(y-cy, 2));

    //System.out.println("Orginal Distance: " + originalDist);
    //System.out.println("New Distance: " + newDist);

    if (newDist > originalDist) {
        return true;
    } else {
        return false;
    }
}

// Find the variables for the original position of the rat
public static int[] findR(String[] enc) {
    for (int i = 0; i < enc.length; i++) {
        for (int j = 0; j < enc[i].length(); j++) {
            if (enc[i].charAt(j) == 'R') {
                int[] coordinates = {i, j};
                //System.out.println(coordinates[0] + "," + coordinates[1]);
                return coordinates;                 
            } else {

            }
        }
    }
    int[] other = {-1, -1};
    return other;
}

// Find the variables for the original position of the rat
public static int[] findC(String[] enc) {
    for (int i = 0; i < enc.length; i++) {
        for (int j = 0; j < enc[i].length(); j++) {
            if (enc[i].charAt(j) == 'C') {
                int[] coordinates = {i, j};
                //System.out.println(coordinates[0] + "," + coordinates[1]);
                return coordinates;                 
            } else {

            }
        }
    }
    int[] other = {-1, -1};
    return other;
}

}

Answer 1

让我们从一个有用的观察开始:

[...] without ever increasing the (Euclidean) distance between itself and the cheese at any point on its path.

基本上，这意味着每当老鼠靠近奶酪时，它就永远不会回到更远的位置。

因此，假设老鼠 x 坐标为 3，奶酪 x 坐标为 5，则老鼠无法“向左走”(即 x = 2)，因为这会导致它比之前距离奶酪更远。

因此，简化问题的一个好方法是找到老鼠可以走的方向。在您的示例中，老鼠位于奶酪的左上方，因此它只能向下或向右移动，否则它会离奶酪更远。当老鼠x与奶酪x匹配时，它将无法再向右或向左移动，y也是如此。

使用您的代码，如果:

r[0] - c[0] = 0 // the rat cannot move on the x any more.
r[1] - c[1] = 0 // the rat cannot move on the y any more.

什么时候

r[0] - c[0] == 0 && r[1] - c[1] == 0

老鼠到达了奶酪!在这种情况下，我们可以增加计数器，因为找到了成功的路由。

现在让我们通过递归来实践这一点。从您发布的代码中，我们从给定的 c(使用 findC(enc) 找到)和给定的 r(使用 findR(enc) 找到)开始

因此递归方法如下所示:

private void findRoutesFrom(int[] r) {
    // what directions can the rat go?
    // if the cheese is on the right of the mouse, xDirection
    // would be +1.
    int xDirection = (int) Math.signum(c[0] - r[0]);
    // if the cheese is below of the mouse, yDirection
    // would be +1.
    int yDirection = (int) Math.signum(c[1] - r[1]);

    // Now, if xDirection is != 0 the rat can attempt to move
    // in that direction, checking if there's not a block
    if(xDirection != 0 && !isBlock(r[0] + xDirection, r[1])) {
        // if it can move in that direction, then use recursion to
        // find all the possible paths form the new position
        findRoutesFrom(new int[]{r[0] + xDirection, r[1]});
    }

    // same goes for yDirection
    if(yDirection != 0 && !isBlock(r[0], r[1] + yDirection)) {
        findRoutesFrom(new int[]{r[0], r[1] + yDirection});
    }

    // if the rat reaches the cheese, increase the successful
    // paths counter
    if(xDirection == 0 && yDirection == 0) {
        count++;
    }
}

就是这样!由于 Eculedean 距离约束，检查方向是否为 != 0 就足够了，因为当方向为 false 时，老鼠就无法再沿该方向移动。