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私はこの問題を試みましたが、何らかの理由でそれが正しく出力されませんでした。一連の文字列が与えられた場合、文字列が1つの "R"（ラット）、1つの "C"（チーズ）、複数の "X"（通過できないブロック） "経路"の任意の点でチーズとの間の（ユークリッド）距離を増やさずに、ラットがチーズに行くために取ることができるルートの数を見つけることです。私のコードで間違って見えるもの再帰を使用してJavaで迷路を解決する
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;
}
}
コード内の特定の例に対する回答は、それが役立つ場合は3となります。 – Steve