java - 加权快速联合

Question

我目前正在拼贴中做算法，我们被要求使用加权快速联合制作一个十六进制游戏，讲师给了我们该项目的大部分代码。但我在这里遇到了问题。`public class Hex Implements BoardGame {

private int[][] board; // 2D Board. 0 - empty, 1 - Player 1, 2 - Player 2 

private int n1, n2; // height and width of board

private WeightedQuickUnionUF wqu; // Union Find data structure to keep track
                                    // of unions and calculate winner

private int currentPlayer; // Current player in the game, initialised to 1

public Hex(int n1, int n2) // create N-by-N grid, with all sites blocked
{
    this.n1 = n1;
    this.n2 = n2;
    currentPlayer = 1;

    // TODO: Create instance of board
    // TODO: Create instance WeightedQuickUnionUF class
    wqu = new WeightedQuickUnionUF(14);
    board = new int[n1][n2];

    for(int i=0; i < n1 ; i++){
        for(int j = 0; j < n2; j++){
            board[i][j] = 0;

        }
    }
}

/*
 * (non-Javadoc)
 * 
 * @see BoardGame#takeTurn(int, int)
 */
@Override
public void takeTurn(int x, int y) {

    if(((x > n1) || (x < 0)) || ((y > n2) || (y < 0)))
    {
        StdOut.println("Wrong");
    } 
    else{

        if(board[x][y] == 0){
            board[x][y] = currentPlayer;
        }
        else{
            StdOut.println("Taken");
        }

    }


    // TODO: check coords are valid
    // TODO: check if location is free and set to player's value(1 or 2).

    // TODO: calculate location and neighbours location in
    // WeightedQuickUnionUF data structure

    // TODO: create unions to neighbour sites in WeightedQuickUnionUF that
    // also contain current players value

    // TODO: if no winner get the next player

}



/*
 * (non-Javadoc)
 * 
 * @see BoardGame#getCurrentPlayer()
 */
@Override
public int getCurrentPlayer() {
    return currentPlayer;
}

public void setCurrentPlayer(int currentPlayer) {
    this.currentPlayer = currentPlayer;
}

/*
 * (non-Javadoc)
 * 
 * @see BoardGame#getBoard()
 */
@Override
public int[][] getBoard() {
    return board;
}

private void nextPlayer() {
    if (currentPlayer == 1)
        currentPlayer = 2;
    else
        currentPlayer = 1;
}

/*
 * (non-Javadoc)
 * 
 * @see BoardGame#isWinner()
 */
@Override
public boolean isWinner() {

    // TODO:check if there is a connection between either side of the board.
    // You can do this by using the 'virtual site' approach in the
    // percolation test.
    return false;
}

/**
 * THIS IS OPTIONAL:
 * Modify the main method if you wish to suit your implementation.
 * This is just an example of a test implementation. 
 * For example you may want to display the board after each turn.
 * @param args
 * 
 */
public static void main(String[] args) {

    BoardGame hexGame = new Hex(4, 4);

    while (!hexGame.isWinner()) {
        System.out.println("It's player " + hexGame.getCurrentPlayer()
                + "'s turn");
        System.out.println("Enter x and y location:");
        int x = StdIn.readInt();
        int y = StdIn.readInt();

        hexGame.takeTurn(x, y);

    }

    System.out.println("It's over. Player " + hexGame.getCurrentPlayer()
            + " wins!");

}

}`

我已经检查过坐标是否有效以及板上的位置是否空闲。但我似乎可以使用 WeightedQuickUnionUF 来查找位置和邻居位置。任何帮助都会很棒，因为我已经尝试了迄今为止我所知道的一切。这是 WeightedQuickUnionUF 类。

public class WeightedQuickUnionUF {

    private int[] id;
    private int[] sz;
    private int count;


public WeightedQuickUnionUF(int N){
    count = N;
    id = new int[N];
    sz = new int[N];
    for(int i = 0 ; i < N; i++){
        id[i] = i;
        sz[i] = i;
        }
}


public int count(){
    return count;
}

public int find(int p){
    while(p != id[p])
        p = id[p];
    return p;
}

public boolean connected(int p, int q ){
    return find(p) == find(q);  
}

public void union(int p, int q){
    int i = find(p);
    int j = find(q);
    if(i == j) return;

    if(sz[i] < sz[j]){id[i] = j; sz[j] += sz[i];}
    else             {id[j] = i; sz[i] += sz[j];}
    count--;
}

public static void main(String[] args) {
    int N = StdIn.readInt();
     WeightedQuickUnionUF uf = new  WeightedQuickUnionUF(N);


     while(!StdIn.isEmpty()){
         int p = StdIn.readInt();
         int q = StdIn.readInt();
         if(uf.connected(p,q)) continue;
         uf.union(p, q);
         StdOut.println(p + " " + q);
     }

     StdOut.println(uf.count() + "components");

    }

}

Answer 1

sz[] 的初始化代码中有一个错误

应该是:

for(int i = 0 ; i < N; i++){
    id[i] = i;
    sz[i] = 1;  // changed to 1 so it indicates the number of nodes for this 'root'
}