因此,我正在用Java编写一个“高峰时间”求解器,它旨在能够解决here配置。但是,即使是该页面上最简单的难题,也会导致求解器无限运行,并最终耗尽内存。我正在使用广度优先搜索来遍历每个板状态产生的所有可能动作(使用HashSet来确保我不会重复自己),并将每个状态映射到到达该状态的动作,以便我可以回溯通过他们以后。

问题是,我尝试了一些自己想出的更简单的难题,并且能够解决这些难题(尽管速度很慢)。

我如何解决这个问题有什么公然的错误?如果需要,我也可以从相关类中放入一些代码,但是我已经对它们进行了彻底的测试,并且我确定问题出在下面的代码中。我的直觉说这与HashSet有关,并确保我不会重复自己(因为Queue的大小经常达到数十万)。

有什么建议么?

    // Start at the original configuration
    queue.add(originalBoard);

    // We add this to our map, but getting here did not require a move, so we use
    // a dummy move as a placeholder move
    previous.put(originalBoard, new Move(-1, -1, "up"));

    // Breadth first search through all possible configurations
    while(!queue.isEmpty()) {
        // Dequeue next board and make sure it is unique
        Board currentBoard = queue.poll();
        if (currentBoard == null)           continue;
        if (seen.contains(currentBoard))    continue;
        seen.add(currentBoard);

        // Check if we've won
        if (currentBoard.hasWon()) {
            System.out.println("We won!");
            currentBoard.renderBoard();
            return solved(currentBoard);
        }

        // Get a list of all possible moves for the current board state
        ArrayList<Move> possibleMoves = currentBoard.allPossibleMoves();

        // Check if one of these moves is the winning move
        for (Move move : possibleMoves) {
            Board newBoard = move.execute(currentBoard);

            // We don't need to enqueue boards we've already seen
            if (seen.contains(newBoard))   continue;
            queue.add(newBoard);

            // Map this board to the move that got it there
            previous.put(newBoard, move);
        }
    }


根据要求,这是我对HashSet的初始化(它们是类级别的变量):

private static HashSet<Board> seen = new HashSet<>();


和我的Board.equals()方法:

@Override
public boolean equals (Object b) {
    Board otherBoard = (Board) b;
    boolean equal = false;
    if (this.M == otherBoard.getM() && this.N == otherBoard.getN()) {
        equal = true;

        // Each board has an ArrayList of Car objects, and boards are only
        // considered equal if they contain the exact same cars
        for (Car car : this.cars) {
            if (otherBoard.getCar(car.getPosition()) == null) {
                equal = false;
            }
        }
    }
    System.out.println(equal);
    return equal;
}

最佳答案

您必须实现Board.hashCode()来覆盖默认的基于Object的版本,以这种方式,根据其合同,任何两个相等的Board对象都具有相同的哈希码。如果您不这样做,那么您的seen集实际上根本不会为您完成任何工作。

在另一个问题上,我怀疑您检查电路板汽车的方式不完全正确。如果按我认为的方式工作,它将认为这两个委员会是平等的:
. =空白
* =汽车的一部分

......
.**.*.
....*.
.*....
.*.**.
......

......
.*..**
.*....
......
.**.*.
....*.

关于java - 高峰时间求解器会遇到任何非平凡难题的无限循环,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/39500162/

10-08 22:11