Question

我为对称的无向图实现Floyd-Warshall算法。目前，我已计算出每个连接点的最佳路径。我的问题是我想保存收取累积重量的索引点，以便以后能够写出路线上的点名。我想将它们保存到列表中，但我不知道哪些索引应写入函数addDrawPointsToList（int a，int b，int [] [] M）。 a和b是我想要保存航路点的点

0 - 同一节点
1 - 节点之间存在连接
X - 无连接= 999重量

代码：

public class FloydWarshallAlg {
static int[][] P;
static List<Integer> lista;
static final int N = 43;
static final int X = 999;

public static void main(String[] args) {

    int M[][] = new int[][]{
        {0, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {1, 0, 1, X, X, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, 1, 0, 1, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, 1, X, 0, X, 1, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, 1, 1, X, X, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, 1, X, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, 1, X, X, 0, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, 1, X, X, 1, 0, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, 1, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, 1, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, X, X, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, X, 0, 1, 1, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, 1, 1, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, X, X, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, X, X, X, 0, 1, X, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, X, X, X, 1, 0, 1, X, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, 1, 1, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X, X, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0, 1, 1},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, 0, X},
        {X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, X, 1, X, 0}};

    lista = new ArrayList<Integer>();
    System.out.println("Main Matrix.");
    printMatrix(M);
    System.out.println("Shortest Path Matrix.");
    printMatrix(FloydAlgo(M));
    addDrawPointsToList(3, 12);
    printList(lista);
}

public static List addDrawPointsToList(int a, int b, int[][] M) {

    return lista;
}

public static int[][] FloydAlgo(int[][] M) {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            for (int k = 0; k < N; k++) {
                // to keep track.;
                if (M[j][i] + M[i][k] < M[j][k]) {
                    M[j][k] = M[j][i] + M[i][k];
                }
            }
        }
    }
    return M;
}

public static void printMatrix(int[][] Matrix) {
    System.out.print("\n\t");
    for (int j = 0; j < N; j++) {
        System.out.print(j + "\t");
    }
    System.out.println();
    for (int j = 0; j < 347; j++) {
        System.out.print("-");
    }
    System.out.println();
    for (int i = 0; i < N; i++) {
        System.out.print(i + " |\t");
        for (int j = 0; j < N; j++) {
            if (Matrix[i][j] == 999) {
                System.out.print("X");
            } else {
                System.out.print(Matrix[i][j]);
            }
            System.out.print("\t");
        }
        System.out.println("");
    }
    System.out.println("\n");
}

public static void printIntArray(int A[]) {
    for (int i = 0; i < N; i++) {
        System.out.print(A[i] + " ");
    }
}

public static void printList(List L) {
    for (int i = 0; i < L.size(); i++) {
        System.out.print(L.get(i) + ", ");
    }
    System.out.println("\nList size: " + L.size());
}}

我觉得这可能是一个需要解决的小问题，但我是一名新手程序员而且我没有看到解决方案。我会很感激任何建议。为我的英语而烦恼;＆lt;

Answer 1

我不知道addDrawPointsToList应该做什么，但一般来说，如果你想从Floyd-Warshall中检索路径，你必须记住你在{{{ 1}}步骤。

如果i-th，那么使用节点M[j][i] + M[i][k] < M[j][k]从j到k的最短路径将经过1, ..., i。因此，最短路径可以在从i到j的最短路径和从i到j的最短路径中分解。让k成为这个中间节点。

A[j][k]

然后获取从if (M[j][i] + M[i][k] < M[j][k]) { A[j][k] = i; M[j][k] = M[j][i] + M[i][k]; }到i的路径：

这是主要的想法，现在你可以编写Java代码来模拟这个。

Floyd-Warshall算法 - 最短路径 - 将路径索引保存到数组

1 个答案: