Java 如何在没有源的情况下对邻接矩阵进行拓扑排序?
因此,我试图找到一种方法来实现这段代码,我可以在不需要输入源代码的情况下对邻接矩阵进行拓扑排序。我试着去查找,但在大多数情况下,其他方法与我的方法相差甚远。有什么帮助吗 代码:Java 如何在没有源的情况下对邻接矩阵进行拓扑排序?,java,arrays,2d,adjacency-matrix,topological-sort,Java,Arrays,2d,Adjacency Matrix,Topological Sort,因此,我试图找到一种方法来实现这段代码,我可以在不需要输入源代码的情况下对邻接矩阵进行拓扑排序。我试着去查找,但在大多数情况下,其他方法与我的方法相差甚远。有什么帮助吗 代码: import java.util.Stack; import java.util.Scanner; import java.util.InputMismatchException; public class Top { private Stack<Integer> stack; publ
import java.util.Stack;
import java.util.Scanner;
import java.util.InputMismatchException;
public class Top
{
private Stack<Integer> stack;
public Top()
{
stack = new Stack<Integer>();
}
public int [] topological(int adjacency_matrix[][], int source) throws NullPointerException
{
int number_of_nodes = adjacency_matrix[source].length - 1;
int[] topological_sort = new int [number_of_nodes + 1];
int pos = 1;
int j ;
int visited[] = new int[number_of_nodes + 1];
int element = source;
int i = source;
visited[source] = 1;
stack.push(source);
while (!stack.isEmpty())
{
element = stack.peek();
while (i <= number_of_nodes)
{
if (adjacency_matrix[element][i] == 1 && visited[i] == 1)
{
if (stack.contains(i))
{
System.out.println("TOPOLOGICAL SORT NOT POSSIBLE");
return null;
}
}
if (adjacency_matrix[element][i] == 1 && visited[i] == 0)
{
stack.push(i);
visited[i] = 1;
element = i;
i = 1;
continue;
}
i++;
}
j = stack.pop();
topological_sort[pos++] = j;
i = ++j;
}
return topological_sort;
}
public static void main(String...arg)
{
int number_no_nodes, source;
Scanner scanner = null;
int topological_sort[] = null;
try
{
System.out.println("Enter the number of nodes in the graph");
scanner = new Scanner(System.in);
number_no_nodes = scanner.nextInt();
int adjacency_matrix[][] = new int[number_no_nodes + 1][number_no_nodes + 1];
System.out.println("Enter the adjacency matrix");
for (int i = 0; i < number_no_nodes; i++)
{
for (int j = 0; j < number_no_nodes; j++)
{
System.out.print("Please enter the value for array["+i+"]["+j+"] (between -10 and 10):");
int val = scanner.nextInt();
if(val>5 || val< -5)
{
System.out.println("Invalid value.");
continue;
}
adjacency_matrix[i][j] = val;
}
}
System.out.println("Enter the source for the graph");
source = scanner.nextInt();
System.out.println("The Topological sort for the graph is given by ");
Top toposort = new Top();
topological_sort = toposort.topological(adjacency_matrix, source);
System.out.println();
for (int i = topological_sort.length - 1; i > 0; i-- )
{
if (topological_sort[i] != 0)
System.out.print(topological_sort[i]+"\t");
}
}
catch(InputMismatchException inputMismatch)
{
System.out.println("Incorrect Input format");
}
catch(NullPointerException nullPointer)
{
}
scanner.close();
}
}
import java.util.Stack;
导入java.util.Scanner;
导入java.util.InputMismatchException;
公务舱头等舱
{
私有堆栈;
公共顶部()
{
堆栈=新堆栈();
}
公共int[]拓扑(int邻接矩阵[],int源)引发NullPointerException
{
节点的整数=邻接矩阵[source]。长度-1;
int[]拓扑_排序=新int[节点数+1];
int pos=1;
int j;
访问的int[]=新int[节点的数量+1];
int元素=源;
int i=源;
访问[来源]=1;
堆栈推送(源);
而(!stack.isEmpty())
{
元素=stack.peek();
而(I5 | | val<-5)
{
System.out.println(“无效值”);
继续;
}
邻接矩阵[i][j]=val;
}
}
System.out.println(“输入图形的源”);
source=scanner.nextInt();
System.out.println(“图的拓扑排序由给出”);
Top toposort=新Top();
拓扑排序=拓扑排序。拓扑(邻接矩阵,源);
System.out.println();
对于(int i=拓扑排序长度-1;i>0;i--)
{
if(拓扑排序[i]!=0)
系统输出打印(拓扑排序[i]+“\t”);
}
}
捕获(输入不匹配异常输入不匹配)
{
System.out.println(“不正确的输入格式”);
}
捕获(NullPointerException nullPointer)
{
}
scanner.close();
}
}
您可以通过遍历找到源,不是吗?事实上,如果您有邻接矩阵,就更容易了。Source是任何一列都是零的。啊!非常感谢。我还在想办法解决这个问题。