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;
}
}
最佳答案
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。
关于java - 在java中使用递归来解决迷宫问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38649044/