java - 求二维数组中最短路径的算法

Question

请帮我实现以下算法。我是初学者，这个问题让我失望。

该算法应该找到从左上角到右下角的最短路径。右下方的元素始终为 0。数组始终为正方形（例如 3x3）。

您只能沿着阵列向下或向右移动。当前位置和int元素，即所谓的跳跃力（比如我们在起点[0][0]，对应的元素是2，那么我们可以向下移动2（D2）-> [2] [0] 或右侧 2 (R2) -> [0] [2])。如果跳跃的力量让我们离开场地（例如，一个 3x3 阵列，我们踩到了 5 号单元格，那么无论如何我们都从两个方向飞离场地），我们需要重新开始并寻找另一种方式。

一种算法应该计算路径/一个人必须跳跃/行走的顺序，以便以尽可能少的跳跃次数到达终点。

算法如何工作的示例（因此，路径将是 [D1, R2, D1] - 向下移动一次，向右移动两次，向下移动一次）

我尝试了不同的方法，现在我在用输入 int[][] 制作隐式图形后正在努力使用 DFS，而我得到的答案不是最短路径。我还听说动态编程可能会有所帮助，但不知道如何实现。

我几乎没有测试的代码包括：

public class Main {

int indexError = 0;
int shortestPathLen = Integer.MAX_VALUE;
List<String> shortestPath = new ArrayList<>();
List<List<String>> allPaths;
List<String> solution = new ArrayList<>();

public List<List<String>> findAllPaths(int[][] map, int D, int R) {
int currentPosition = map[D][R]; //D - down, R - right
 if (currentPosition == 0) {
allPaths.add(solution);
return allPaths;
}
if (D + currentPosition <= indexError) {
solution.add("D" + currentPosition);
findAllPaths(map, D+currentPosition, R);
}
if (R + currentPosition <= indexError) {
solution.add("R" + currentPosition);
findAllPaths(map, D, R+currentPosition);
}
solution = new ArrayList<>();
return allPaths;
}  

public List<String> findPath(int[][] map) {
indexError = map[0].length - 1;
shortestPathLen = Integer.MAX_VALUE;
allPaths = new ArrayList<>();

List<List<String>> l = findAllPaths(map, 0, 0);
for (List<String> path : l) {
if (path.size() < shortestPathLen) {
    shortestPathLen = path.size();
    shortestPath = path;
}
}
return shortestPath;
}

public static void main(String[] args) {
Main main = new Main();
// int len = 3;
// int[] array =   {1, 2, 2,
//                 2, 10, 1,
//                 3, 2, 0}; // from example
int len = 9;
int[] array =
            {1, 10, 20, 1, 2, 2, 1, 2, 2,
            1, 10, 1, 10, 2, 2, 1, 2, 2,
            1, 10, 1, 1, 20, 2, 1, 2, 2,
            2, 1, 10, 1, 1, 20, 1, 2, 2,
            1, 2, 2, 10, 1, 1, 10, 2, 2,
            2, 1, 1, 1, 10, 1, 1, 20, 2,
            1, 2, 2, 1, 2, 10, 1, 1, 20,
            1, 1, 1, 1, 2, 2, 10, 1, 1,
            1, 2, 1, 1, 2, 2, 1, 1, 0};
int[][] map = new int[len][len];
int k = 0;
for (int i = 0; i < len; i++) {
   for (int j = 0; j < len; j++) {
    map[i][j] = array[k];
    k++;
  }
}
List result = main.findPath(map);
System.out.println("\n" + result + ", " + result.size() + " jumps");
// = must be [D1, D1, D1, D2, R2, D1, R2, D2, R2, R1, R1], 11 jumps
}}

标签： javaalgorithmdynamic-programmingshortest-path

Answer 1

使用此代码（A*-Search），您将不会获得最佳解决方案，但几乎不会。

//= D1, D1, D1, D2, D2, D1, R1, R2, R1, R2, R1, R1, 12 jumps

如果您可以更改distanceToGoal() - 方法以更好地估计到目标的真实距离，您将获得最佳解决方案。

public class TestClass {
    private static class Position implements Comparable<Position>{
        private final int[][] map;
        private int x;
        private int y;
        List<String> path = new ArrayList<>();

        /**
         * Some Constructors
        */
        public Position(int[][] map) {
           this(map, 0, 0);
        }

        public Position(Position parent, Consumer<Position> consumer, String action) {
            this(parent.map, parent.x, parent.y);
            consumer.accept(this);

            this.path.addAll(parent.path);
            this.path.add(action);
        }

        private Position(int[][] map, int x, int y) {
            this.map = map;
            this.x = x;
            this.y = y;
        }

        /**
         * Returns Jumpforce of current position
         */
        public int getJumpForce() {
            return this.map[this.y][this.x];
        }

        /**
         * Returns steps taken till current position
         */
        public int steps() {
            return CollectionUtils.size(this.path);
        }

        /**
         * Method calculates the estimated way to the goal. In this case the mannhatten distance
         */
        private int distanceToGoal() {
            var height = ArrayUtils.getLength(this.map);
            var width = ArrayUtils.getLength(this.map);
            return (height - (y+1)) + (width - (x+1));
        }

        /**
         * Sum of steps taken and the estimated distance till the goal
         */
        private int totalCosts() {
           return this.steps() + this.distanceToGoal();
        }

        public boolean isGoal() {
            return this.map[this.y][this.x] == 0;
        }
    
        /**
         * Returns a list of all successor states
         */
        public List<Position> successors() {
            var s = new ArrayList<Position>();
            if (ArrayUtils.getLength(this.map) > (this.y + this.getJumpForce())) {
                s.add(new Position(this,
                        p -> p.y += this.getJumpForce(),
                        String.format("D%d", this.getJumpForce())));
            }
            if (ArrayUtils.getLength(map[y]) > (x + this.getJumpForce())) {
                s.add(new Position(this,
                        p -> p.x += this.getJumpForce(),
                        String.format("R%d", this.getJumpForce())));
            }
            return s;
        }

    
        @Override
        public boolean equals(Object o) {
            if (this == o) return true;

            if (o == null || getClass() != o.getClass()) return false;

            Position position = (Position) o;

            return new EqualsBuilder()
                    .append(x, position.x)
                    .append(y, position.y)
                    .isEquals();
        }

        @Override
        public int hashCode() {
            return new HashCodeBuilder(17, 37)
                    .append(x)
                    .append(y)
                    .toHashCode();
        }

        @Override
        public int compareTo(Position o) {
            return Comparator.comparing(Position::totalCosts).compare(this, o);
        }

        @Override
        public String toString() {
            return String.join(", ", path);
        }
    }

    public static Position findShortestPath(int[][] map) {
        var visited = new HashSet<Position>();
        var fringe = new PriorityQueue<Position>(){{
            add(new Position(map));
        }};
    
        // As long as there is a position to check
        while (CollectionUtils.isNotEmpty(fringe)) {
            // Get that position
            var position = fringe.poll();
            //If we didn't look already at this position
            if (!visited.contains(position)) {
                // Check if its the goal
                if (position.isGoal()) {
                    return position;
                } else {
                    //Mark position as visited
                    visited.add(position);
                    //Add all successors to be checked
                    fringe.addAll(position.successors());
                }
            }
        }
    
        // If no solution is found
        return null;
    }

    public static void main(String[] args) {
        // int len = 3;
        // int[] array =   {1, 2, 2,
        //                 2, 10, 1,
        //                 3, 2, 0}; // from example
        int[][] map = { {1, 10, 20, 1, 2, 2, 1, 2, 2},
                        {1, 10, 1, 10, 2, 2, 1, 2, 2},
                        {1, 10, 1, 1, 20, 2, 1, 2, 2},
                        {2, 1, 10, 1, 1, 20, 1, 2, 2},
                        {1, 2, 2, 10, 1, 1, 10, 2, 2},
                        {2, 1, 1, 1, 10, 1, 1, 20, 2},
                        {1, 2, 2, 1, 2, 10, 1, 1, 20},
                        {1, 1, 1, 1, 2, 2, 10, 1, 1},
                        {1, 2, 1, 1, 2, 2, 1, 1, 0} };

        Position position = findShortestPath(map);
        if (Objects.nonNull(position)) {
            System.out.printf("%s, %d jumps%n", position, position.steps());
        } else {
            System.out.println("No solution found");
        }
        // = must be [D1, D1, D1, D2, R2, D1, R2, D2, R2, R1, R1], 11 jumps
    }

}