[Daily Blue Bridge] 46.17 Provincial Competition Java Group Real Question "Magic Cube State"

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Title: Rubik's Cube Status

The second-order Rubik's Cube is a Rubik's Cube with only 2 layers, consisting of only 8 small pieces.

As shown in figure pl.png.

Xiao Ming is very naughty, he only likes 3 colors, so he repainted the second-order Rubik's Cube at home, as follows:

Front: orange

Right: Green

Top: yellow

Left: green

Below: orange

Back: yellow

Please calculate how many different states there are after such a Rubik's Cube is disrupted.

If the two states are the same after the overall rotation of the Rubik's Cube, the colors of all sides are the same, it is considered the same state.

Please submit an integer representing the number of states, and do not fill in any redundant content or explanatory text.

Problem-solving ideas:

This question mainly examines the candidate’s spatial thinking and logical imagination. The solution is mainly to simulate each magic cube. The magic cube in the question consists of 8 small cubes, and each cube has 6 faces. Use a string to divide this The 6 faces are represented, and then the rotation process of each face is simulated, the changes of each face are recorded during the rotation, and the corresponding character string is recorded, and the character sequence is compared to see if the character sequence is the same, and then the different rotations can be obtained. State now.

Answer source code:

import java.util.HashSet;
public class Year2017_Bt4 {
  static char[][] start = {"oybbgb".toCharArray(),
      "oygbbb".toCharArray(),
      "bygbby".toCharArray(),
      "bybbgy".toCharArray(),
      "obbogb".toCharArray(),
      "obgobb".toCharArray(),
      "bbgoby".toCharArray(),
      "bbbogy".toCharArray()};
  static char[][][] q = new char[2000000][8][6];
  static HashSet<String> all_state = new HashSet<String>();
  static int  front, tail;
  static String to_string(char[][] a) {
    String ans = "";
    for (int i = 0; i < 8; ++i) {
      ans += new String(a[i]);
    }
    return ans; 
  }
 
  private static void swap(char[] a, int i, int j) {
    char t = a[i];
    a[i] = a[j];
    a[j] = t;
  }
 
  private static void swap(char[][] a, int i, int j) {
    char[] t = a[i];
    a[i] = a[j];
    a[j] = t;
  }
 
  //上层的块的旋转，面的相对位置调换
  static void ucell(char[] a) {
    swap(a, 0, 2);
    swap(a, 2, 5);
    swap(a, 5, 4);
  }
 
 
  //上层顺时针旋转
  static void u(char[][] s) {
    ucell(s[0]);
    ucell(s[1]);
    ucell(s[2]);
    ucell(s[3]);
//    块的相对位置调换
    swap(s, 1, 0);
    swap(s, 2, 1);
    swap(s, 3, 2);
 
  }
 
  //右层旋转是面的位置变化
  static void rcell(char[] a) {
    swap(a, 1, 0);
    swap(a, 0, 3);
    swap(a, 3, 5);
  }
 
  static void r(char[][] s)//魔方右层顺时针转
  {
    rcell(s[1]);
    rcell(s[2]);
    rcell(s[6]);
    rcell(s[5]);
//    块的位置变化
    swap(s, 2, 1);
    swap(s, 5, 1);
    swap(s, 6, 5);
  }
 
  static void fcell(char[] a) {
    swap(a, 2, 1);
    swap(a, 1, 4);
    swap(a, 4, 3);
  }
 
  static void f(char[][] s)//前面一层 顺时针转
  {
    fcell(s[0]);
    fcell(s[1]);
    fcell(s[4]);
    fcell(s[5]);
    swap(s, 1, 5);
    swap(s, 0, 1);
    swap(s, 4, 0);
  }
 
  static void uwhole(char[][] s)//整个魔方从顶部看 顺时针转 用于判重
  {
    u(s);//上层旋转
//    下层旋转
    ucell(s[4]);
    ucell(s[5]);
    ucell(s[6]);
    ucell(s[7]);
//    完成自旋后，块的位置变动
    swap(s, 5, 4);
    swap(s, 6, 5);
    swap(s, 7, 6);
  }
 
  static void fwhole(char[][] s)//整个魔方从前面看 顺时针转 用于判重
  {
    f(s);
    fcell(s[2]);
    fcell(s[6]);
    fcell(s[7]);
    fcell(s[3]);
    swap(s, 2, 6);
    swap(s, 3, 2);
    swap(s, 7, 3);
  }
 
  static void rwhole(char[][] s)//整个魔方从右边看 顺时针转 用于判重
  {
    r(s);
    rcell(s[0]);
    rcell(s[3]);
    rcell(s[4]);
    rcell(s[7]);
    swap(s, 3, 7);
    swap(s, 0, 3);
    swap(s, 4, 0);
  }
 
 
  static boolean try_insert(char[][] s) {
    char[][] k = new char[8][6];
    memcpy(k, s);
    for (int i = 0; i < 4; i++) {
      fwhole(k);
      for (int j = 0; j < 4; j++) {
        uwhole(k);
        for (int q = 0; q < 4; q++) {
          rwhole(k);
          if (all_state.contains(to_string(k))) {
            return false;
          }
        }
      }
    }
    all_state.add(to_string(k));
    return true;
 
  }
 
  private static void memcpy(char[][] k, char[][] s) {
    for (int i = 0; i < 8; i++) {
      for (int j = 0; j < 6; j++) {
        k[i][j] = s[i][j];
      }
    }
  }
 
  static void solve() {
    front = 0;
    tail = 1;
    all_state.add(to_string(start));
    memcpy(q[front], start);//填充q[0]，相当于第一个状态入队列
    while (front < tail) {
        /*将其所有变形，尝试加入set中*/
      memcpy(q[tail], q[front]);//拷贝到tail
      u(q[tail]);//上层顺时针旋转
      if (try_insert(q[tail])) {
        tail++;//扩展队列
      }
      memcpy(q[tail], q[front]);//拷贝到tail
      r(q[tail]);//右层顺时针旋转
      if (try_insert(q[tail])) {
        tail++;//扩展队列
      }
      memcpy(q[tail], q[front]);//拷贝到tail
      f(q[tail]);//前顺时针旋转
      if (try_insert(q[tail])) {
        tail++;//扩展队列
      }
      front++;//弹出队首
//        cout << front << " " << tail << endl;
    }
 
    System.out.println(front);
  }
 
  public static void main(String[] args) {
    solve();
  }
}

Sample output:

There is insufficient or those areas of improvement, but also hope that the message put forward a small partner, learn together!

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