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我需要通过生成所有排列来检查图是否同构。我正在使用这个排列类，现在我需要创建一个图形类，将图形表示为 2D 布尔数组。一个例子是用户输入 2 个图表作为字符串

Ex."0-1 0-2 1-2 1-3 2-3" and "1-3 2-0 0-3 1-2 1-0"

现在在我的 constrcutor 中，我必须将其分解并将其放入 2D 布尔数组中。我该怎么做呢？

用户还可以输入更复杂的内容，例如“0-1 1-2 2-3 3-0 0-4 0-11 1-5 1-6 2-7 2-8 3-9 3-10 4-5 6-7 8-9 10-11 4-7 4-8 5-9 5-10 6-9 6-10 7-11 8-11”和“0-1 1-2 2-3 3-0 0- 4 0-7 1-4 1-5 2-5 2-6 3-6 3-7 4-8 4-9 5-9 5-10 6-10 6-11 7-8 7-11 8-9 9 -10 10-11 11-8"

public class PermutationGenerator
{
    // private data

    private int[] perm;
    private boolean first;

    // constructor

    public PermutationGenerator (int n)
    {
        perm = new int [n];
        first = true;
    }



    public int[] next ()
    {
        int n = perm.length;

        // starting permutation: 0 1 2 3 ... n-1

        if (first)
        {
            first = false;
            for (int i = 0 ; i < n ; i++)
                perm [i] = i;
            return perm;
        }

        // construct the next permutation
        // find largest k so that perm[k] < perm[k+1]; if none, finish

        int i, j, k, l;

        for (k = n - 2 ; k >= 0 && perm [k] >= perm [k + 1] ; k--)
            ;
        if (k < 0)
            return null; // no more

        // find largest l so that perm[k] < perm[l]

        for (l = n - 1 ; l >= 0 && perm [k] >= perm [l] ; l--)
            ;

        // swap perm[k] and perm[l]

        swap (perm, k, l);

        // reverse perm[k+1]...perm[n-1]

        for (i = k + 1, j = n - 1 ; i < j ; i++, j--)
            swap (perm, i, j);

        return perm;
    }


    // swap a[i] and a[j]

    private static void swap (int a[], int i, int j)
    {
        int temp = a [i];
        a [i] = a [j];
        a [j] = temp;
    }
}
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1 回答 1

3

You have to create a matrix to represent your graph, as shown in this picture:

enter image description here

You can start by prompting the user for two nodes that are connected, e.g., 1 and 2. Then you would do

matrix[1][2] = true;

if you are taking into account the direction, which means that 1 - 2 is different than 2 - 1. Otherwise (a graph non-oriented), you would do

matrix[1][2] = true;
matrix[2][1] = true;

Do not forget to initialize the matrix with false, meaning (vertices not connected). Every time you want to check if vertices X and Y are directly connected you just have to access the position (X,Y) of the matrix and check if it is true.

If your graph is a weighted one, you need to have an additional matrix to hold the weights. Another solution is to have just one matrix of integers where -1 means not connected, and N > 0, means that a given X and Y are connected with a weight N.

About the user input:

"0-1 1-2 2-3 3-0 0-4 0-11 1-5 1-6 2-7 2-8 3-9 3-10 4-5 6-7 8-9 10-11 4-7 4-8 5-9 5-10 6-9 6-10 7-11 8-11"

among others, you can use the Scanner class:

String input;
Scanner in = new Scanner(System.in);

// Reads a single line from the console 
// and stores into name variable
input = in.nextLine();

Afterwards, you have to parser your input; let us assume a fixed format that will hold the graph vertices, for example:

  String input =  "0-1 0-2 1-2 1-3 2-3";
  int x,y;

  for(int i = 0; i < input.length(); i+=4)
  {
       x = Character.getNumericValue(input.charAt(i));   // first vertex 
       y = Character.getNumericValue(input.charAt(i+2)); // second vertice 
       matrix_graph[x][y] = true;
       matrix_graph[y][x] = true; // if the graph is not oriented.
  }
于 2012-12-04T00:31:47.417 回答