我正在尝试进行深度优先遍历。我不知道我是否接近。现在正在打印1 3 4 5.应该打印1 2 4 7 3 5 6.感谢任何帮助或建议。谢谢。 :)
类别:
public class myGraphs {
Stack<Integer> st;
int vFirst;
int[][] adjMatrix;
int[] isVisited = new int[7];
public myGraphs(int[][] Matrix) {
this.adjMatrix = Matrix;
st = new Stack<Integer>();
int i;
int[] node = {1, 2, 3, 4, 5, 6, 7};
int firstNode = node[0];
for (i = 1; i < node.length - 1; i++) {
depthFirst(firstNode, node[i]);
}
}
public void depthFirst(int vFirst, int n) {
int v, i;
st.push(vFirst);
while (!st.isEmpty()) {
v = st.pop();
if (isVisited[v]==0) {
System.out.print("\n"+v);
isVisited[v]=1;
}
for ( i=1;i<=n;i++) {
if ((adjMatrix[v][i] == 1) && (isVisited[i] == 0)) {
st.push(v);
isVisited[i]=1;
System.out.print(" " + i);
v = i;
}
}
}
}
//
public static void main(String[] args) {
// 1 2 3 4 5 6 7
int[][] adjMatrix = { {0, 1, 1, 0, 0, 0, 0},
{1, 0, 0, 1, 1, 1, 0},
{1, 0, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0 ,0},
{0, 0, 1, 1, 1, 0, 0} };
new myGraphs(adjMatrix);
}
}
答案 0 :(得分:4)
如果您正在查看Depth First Traversal,那么以下是您应该进行的代码更改
1)首先将节点数组声明为int[] node = {0, 1, 2, 3, 4, 5, 6}
。这应该是为了避免数组索引启动(为0)和节点起始编号(为1)。现在我们假设您的节点1的新名称为0,节点2为1 ......节点7为6。
2)而不是做
for (i = 1; i < node.length-1; i++){
depthFirst(firstNode, node[i]);
}
在myGraphs中: depthFirst(firstNode,7);
3)在depthFirst而不是for ( i=1;i<=n;i++)
中使用for ( i=0;i<n;i++)
在函数depthFirst中执行System.out.println时,在数字中添加一个,因为0代表节点1,1代表节点2,依此类推。 / p>
以下是我修改的全功能代码:
import java.util.Stack;
public class DFS {
Stack<Integer> st;
int vFirst;
int[][] adjMatrix;
int[] isVisited = new int[7];
/**
* @param args
*/
public static void main(String[] args) {
int[][] adjMatrix = { {0, 1, 1, 0, 0, 0, 0},
{1, 0, 0, 1, 1, 1, 0},
{1, 0, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0 ,0},
{0, 0, 1, 1, 1, 0, 0} };
new DFS(adjMatrix);
}
public DFS(int[][] Matrix) {
this.adjMatrix = Matrix;
st = new Stack<Integer>();
int i;
int[] node = {0, 1, 2, 3, 4, 5, 6};
int firstNode = node[0];
depthFirst(firstNode, 7);
}
public void depthFirst(int vFirst,int n)
{
int v,i;
st.push(vFirst);
while(!st.isEmpty())
{
v = st.pop();
if(isVisited[v]==0)
{
System.out.print("\n"+(v+1));
isVisited[v]=1;
}
for ( i=0;i<n;i++)
{
if((adjMatrix[v][i] == 1) && (isVisited[i] == 0))
{
st.push(v);
isVisited[i]=1;
System.out.print(" " + (i+1));
v = i;
}
}
}
}}
答案 1 :(得分:0)
C#中的工作/测试解决方案,如果有人在寻找它。
using System;
using System.Collections.Generic;
namespace GraphAdjMatrixDemo
{
public class Program
{
public static void Main(string[] args)
{
// 0 1 2 3 4 5 6
int[,] matrix = { {0, 1, 1, 0, 0, 0, 0},
{1, 0, 0, 1, 1, 1, 0},
{1, 0, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0, 1},
{0, 1, 0, 0, 0, 0 ,0},
{0, 0, 1, 1, 1, 0, 0} };
bool[] visitMatrix = new bool[matrix.GetLength(0)];
Program ghDemo = new Program();
for (int lpRCnt = 0; lpRCnt < matrix.GetLength(0); lpRCnt++)
{
for (int lpCCnt = 0; lpCCnt < matrix.GetLength(1); lpCCnt++)
{
Console.Write(string.Format(" {0} ", matrix[lpRCnt, lpCCnt]));
}
Console.WriteLine();
}
Console.Write("\nDFS Recursive : ");
ghDemo.DftRecursive(matrix, visitMatrix, 0);
Console.Write("\nDFS Iterative : ");
ghDemo.DftIterative(matrix, 0);
Console.Read();
}
//====================================================================================================================================
public void DftRecursive(int[,] srcMatrix, bool[] visitMatrix, int vertex)
{
visitMatrix[vertex] = true;
Console.Write(vertex + 1 + " ");
for (int neighbour = 0; neighbour < srcMatrix.GetLength(0); neighbour++)
{
if (visitMatrix[neighbour] == false && srcMatrix[vertex, neighbour] == 1)
{
DftRecursive(srcMatrix, visitMatrix, neighbour);
}
}
}
public void DftIterative(int[,] srcMatrix, int srcVertex)
{
bool[] visited = new bool[srcMatrix.GetLength(0)];
Stack<int> vertexStack = new Stack<int>();
vertexStack.Push(srcVertex);
while (vertexStack.Count > 0)
{
int vertex = vertexStack.Pop();
if (visited[vertex] == true)
continue;
Console.Write(vertex + 1 + " ");
visited[vertex] = true;
for (int neighbour = 0; neighbour < srcMatrix.GetLength(0); neighbour++)
//for (int neighbour = srcMatrix.GetLength(0) - 1; neighbour >= 0; neighbour--)// To make same as recursive
{
if (srcMatrix[vertex, neighbour] == 1 && visited[neighbour] == false)
{
vertexStack.Push(neighbour);
}
}
}
}
}
}
为了使迭代的显示顺序与递归相同,我们需要以相反的顺序推送邻居来进行堆栈。从Amit回答here
这个逻辑