邻接表代码JAVA 邻接表c++实现
java中如何把图用邻接表表示出来
package my.graph;
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import java.util.ArrayList;
import java.util.Iterator;
import my.queue.*;
import my.stack.StackX;
/**
* 邻接表表示
* @author xiayi
*
*/
public class Graph {
private int MAX_VERTS = 20;
private Vertex vertexList[];
private boolean is = false;//是否为有向图
private int nVerts = 0;
private StackX stackX;
private Vertex dfs[];
private Vertex bfs[];
private Queue queue;
public Graph(){
vertexList = new Vertex[MAX_VERTS];
dfs = new Vertex[MAX_VERTS];
bfs = new Vertex[MAX_VERTS];
}
public Graph(int n){
vertexList = new Vertex[n];
dfs = new Vertex[n];
bfs = new Vertex[n];
}
public Graph(int n, boolean is){
this.is = is;
vertexList = new Vertex[n];
dfs = new Vertex[n];
bfs = new Vertex[n];
}
//////////////////////////////////////////////
public boolean isIs() {
return is;
}
public void setIs(boolean is) {
this.is = is;
}
public Vertex[] getVertexList() {
return vertexList;
}
public Vertex[] getDfs() {
return dfs;
}
public Vertex[] getBfs() {
return bfs;
}
////////////////////////////////////////////////////
/**
* 添加顶点
*/
public void addVertex(Vertex vertex){
vertex.setIndex(nVerts);
vertexList[nVerts] = vertex;
nVerts++;
}
/**
* 添加边
*/
public void addEdge(int start, int end){
vertexList.addAdj(vertexList);
if (!is) {vertexList.addAdj(vertexList);}
}
/**
* 返回节点个数
* @return
*/
public int getVertsCount(){
return vertexList.length;
}
/**
* 深度优先迭代器
* @return
*/
public Iterator dfsIterator(){
dfs();
return new DfsIterator();
}
/**
* 广度优先迭代器
* @return
*/
public Iterator bfsIterator(){
bfs();
return new BfsIterator();
}
////////////////////////////////////////////////////////
public void displayGraph(){
ArrayListVertex next = null;
for (int i = 0; i vertexList.length; i++) {
printVertx(vertexList[i]);
}
}
public void printVertx(Vertex vertex){
ArrayListVertex next = vertex.getAdj();
if(next == null){ System.out.println(vertex.toString()+" 无连接点");}
else{
System.out.print(vertex.toString()+"有邻接点:");
for (int i = 0; i next.size(); i++) {
System.out.print("顶点"+next.get(i).label+", ");
}
System.out.println();
}
}
///////////////////////////////////////////////////////////
public void dfs(){
stackX = new StackX(MAX_VERTS);
vertexList[0].isVisted = true;
dfs[0] = vertexList[0];
stackX.push(vertexList[0]);
int dfsIndex = 0;
Vertex vertex;
while(!stackX.isEmpty()){
vertex = getAdjVertex((Vertex)stackX.peek());
if(vertex == null){
stackX.pop();
}else{
vertex.isVisted = true;
dfs[++dfsIndex]=vertex;
stackX.push(vertex);
}
}
for (int i = 0; i getVertsCount(); i++) {
vertexList[i].isVisted = false;
}
}
public void bfs() {
queue = new Queue(MAX_VERTS);
vertexList[0].isVisted = true;
bfs[0] = vertexList[0];
queue.insert(vertexList[0]);
int bfsIndex = 0;
Vertex vertex;
while(!queue.isEmpty()){
Vertex vertex2 = (Vertex)queue.remove();
while((vertex = getAdjVertex(vertex2))!=null){
vertex.isVisted = true;
bfs[++bfsIndex] = vertex;
queue.insert(vertex);
}
}
for (int i = 0; i getVertsCount(); i++) {
vertexList[i].isVisted = false;
}
}
/**
* 得到一个邻接点
* @param vertex
* @return
*/
public Vertex getAdjVertex(Vertex vertex){
ArrayListVertex adjVertexs = vertex.getAdj();
for (int i = 0; i adjVertexs.size(); i++) {
if(!adjVertexs.get(i).isVisted){
return adjVertexs.get(i);
}
}
return null;
}
/////////////////////////////////////////////////////////////
private abstract class GraphIterator implements Iterator{
int count = 0;
public GraphIterator(){
}
public boolean hasNext() {
return count != getVertsCount()-1;
}
public Object next() {
// TODO Auto-generated method stub
return null;
}
public void remove() {
// TODO Auto-generated method stub
}
}
//深度优先迭代
private class DfsIterator extends GraphIterator{
public DfsIterator(){
super();
}
public Vertex next() {
return dfs[count++];
}
}
//广度优先迭代
private class BfsIterator extends GraphIterator{
public BfsIterator(){
super();
}
public Object next() {
return bfs[count++];
}
}
/////////////////////////////////////////////////////////
public static void main(String[] args) {
int nVerts = 10;
int c = 'A'-1;
Vertex vertex;
Graph myGraph = new Graph(nVerts, false);
for (int i = 0; i nVerts; i++) {
c++;
vertex = new Vertex((char)(c));
myGraph.addVertex(vertex);
}
myGraph.addEdge(0, 1);
myGraph.addEdge(0, 4);
myGraph.addEdge(1, 2);
myGraph.addEdge(2, 3);
myGraph.addEdge(4, 5);
myGraph.addEdge(4, 6);
myGraph.addEdge(5, 8);
myGraph.addEdge(6, 7);
myGraph.addEdge(7, 8);
myGraph.addEdge(8, 9);
System.out.println("深度优先迭代遍历:");
for (Iterator iterator = myGraph.dfsIterator(); iterator.hasNext();) {
vertex = (Vertex) iterator.next();
System.out.println(vertex.toString());
}
System.out.println("/n广度优先迭代遍历:");
for (Iterator iterator = myGraph.bfsIterator(); iterator.hasNext();) {
vertex = (Vertex) iterator.next();
System.out.println(vertex.toString());
}
}
}
class Vertex{
public char label;
public boolean isVisted;
public int index;
private ArrayListVertex next = null;
public Vertex(char lab) // constructor
{
label = lab;
isVisted = false;
}
//为节点添加邻接点
public void addAdj(Vertex ver){
if(next == null) next = new ArrayListVertex();
next.add(ver);
}
public ArrayListVertex getAdj(){
return next;
}
public void setIndex(int index){
this.index = index;
}
public String toString(){
return "顶点 "+label+",下标:"+index+".";
}
}
代码来自:
求代码,java实验,题目如图
import java.util.Scanner;
import java.util.Stack;
public class DFS
{
// 存储节点信息
private char[] vertices;
// 存储边信息(邻接矩阵)
private int[][] arcs;
// 图的节点数
private int vexnum;
// 记录节点是否已被遍历
private boolean[] visited;
// 初始化
public DFS(int n)
{
vexnum = n;
vertices = new char[n];
arcs = new int[n][n];
visited = new boolean[n];
for(int i = 0; i vexnum; i++)
{
for(int j = 0; j vexnum; j++)
{
arcs[i][j] = 0;
}
}
}
// 添加边(无向图)
public void addEdge(int i, int j)
{
// 边的头尾不能为同一节点
if(i == j)
return;
arcs[i - 1][j - 1] = 1;
arcs[j - 1][i - 1] = 1;
}
// 设置节点集
public void setVertices(char[] vertices)
{
this.vertices = vertices;
}
// 设置节点访问标记
public void setVisited(boolean[] visited)
{
this.visited = visited;
}
// 打印遍历节点
public void visit(int i)
{
System.out.print(vertices[i] + " ");
}
// 从第i个节点开始深度优先遍历
private void traverse(int i)
{
// 标记第i个节点已遍历
visited[i] = true;
// 打印当前遍历的节点
visit(i);
// 遍历邻接矩阵中第i个节点的直接联通关系
for(int j = 0; j vexnum; j++)
{
// 目标节点与当前节点直接联通,并且该节点还没有被访问,递归
if(arcs[i][j] == 1 visited[j] == false)
{
traverse(j);
}
}
}
// 图的深度优先遍历(递归)
public void DFSTraverse(int start)
{
// 初始化节点遍历标记
for(int i = 0; i vexnum; i++)
{
visited[i] = false;
}
// 从没有被遍历的节点开始深度遍历
for(int i = start - 1; i vexnum; i++)
{
if(visited[i] == false)
{
// 若是连通图,只会执行一次
traverse(i);
}
}
}
// 图的深度优先遍历(非递归)
public void DFSTraverse2(int start)
{
// 初始化节点遍历标记
for(int i = 0; i vexnum; i++)
{
visited[i] = false;
}
StackInteger s = new StackInteger();
for(int i = start - 1; i vexnum; i++)
{
if(!visited[i])
{
// 连通子图起始节点
s.add(i);
do
{
// 出栈
int curr = s.pop();
// 如果该节点还没有被遍历,则遍历该节点并将子节点入栈
if(visited[curr] == false)
{
// 遍历并打印
visit(curr);
visited[curr] = true;
// 没遍历的子节点入栈
for(int j = vexnum - 1; j = 0; j--)
{
if(arcs[curr][j] == 1 visited[j] == false)
{
s.add(j);
}
}
}
} while(!s.isEmpty());
}
}
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
int N, M, S;
while(true)
{
System.out.println("输入N M S,分别表示图G的结点数,边数,搜索的起点:");
String line = sc.nextLine();
if(!line.matches("^\\s*([1-9]\\d?|100)(\\s+([1-9]\\d?|100)){2}\\s*$"))
{
System.out.print("输入错误,");
continue;
}
String[] arr = line.trim().split("\\s+");
N = Integer.parseInt(arr[0]);
M = Integer.parseInt(arr[1]);
S = Integer.parseInt(arr[2]);
break;
}
DFS g = new DFS(N);
char[] vertices = new char[N];
for(int i = 0; i N; i++)
{
vertices[i] = (i + 1 + "").charAt(0);
}
g.setVertices(vertices);
for(int m = 0; m M; m++)
{
System.out.println("输入图G的第" + (m + 1) + "条边,格式为“i j”,其中i,j为结点编号(范围是1~N)");
String line = sc.nextLine();
if(!line.matches("^\\s*([1-9]\\d?|100)\\s+([1-9]\\d?|100)\\s*$"))
{
System.out.print("输入错误,");
m--;
continue;
}
String[] arr = line.trim().split("\\s+");
int i = Integer.parseInt(arr[0]);
int j = Integer.parseInt(arr[1]);
g.addEdge(i, j);
}
sc.close();
System.out.print("深度优先遍历(递归):");
g.DFSTraverse(S);
System.out.println();
System.out.print("深度优先遍历(非递归):");
g.DFSTraverse2(S);
}
}
BFS求源代码及思路?
1、算法用途:
是一种图像搜索演算法。用于遍历图中的节点,有些类似于树的深度优先遍历。这里唯一的问题是,与树不同,图形可能包含循环,因此我们可能会再次来到同一节点。
2、主要思想:
主要借助一个队列、一个布尔类型数组、邻接矩阵完成(判断一个点是否查看过,用于避免重复到达同一个点,造成死循环等),先将各点以及各点的关系存入邻接矩阵。
再从第一个点开始,将一个点存入队列,然后在邻接表中找到他的相邻点,存入队列,每次pop出队列头部并将其打印出来(文字有些抽象,实际过程很简单),整个过程有点像往水中投入石子水花散开。
(邻接表是表示了图中与每一个顶点相邻的边集的集合,这里的集合指的是无序集)
3、代码(java):
(以上图为例的代码)
1 import java.util.*; 2 3 //This class represents a directed graph using adjacency list
4 //representation 5 class Graph1 { 6 private static int V; // No. of vertices 7 private LinkedListInteger a Lists 8 9 // Constructor10 Graph1(int v) {11 V = v;12 adj = new LinkedList[v];13 for (int i = 0; i v; ++i)14 adj[i] = new LinkedList();15 }16 17 // Function to add an edge into the graph18 void addEdge(int v, int w) {19 adj[v].add(w);20 }21 22 // prints BFS traversal from a given source s23 public void BFS() {24 // Mark all the vertices as not visited(By default25 // set as false)26 boolean visited[] = new boolean[V];27 // Create a queue for BFS28 LinkedListInteger queue = new LinkedListInteger();29 30 for (int i = 0; i V; i++) {31 if (!visited[i]) {32 BFSUtil(i, visited, queue);33 }34 }35 }36 37 public void BFSUtil(int s, boolean visited[], LinkedListInteger queue) {38 // Mark the current node as visited and enqueue it39 visited[s] = true;40 queue.add(s);41 42 while (queue.size() != 0) {43 // Dequeue a vertex from queue and print it44 s = queue.poll();45 System.out.print(s + " ");46 47 // Get all adjacent vertices of the dequeued vertex s48 // If a adjacent has not been visited, then mark it49 // visited and enqueue it50 IteratorInteger i = adj[s].listIterator();51 while (i.hasNext()) {52 int n = i.next();53 if (!visited[n]) {54 visited[n] = true;55 queue.add(n);56 }57 }58 }59 }60 61 // Driver method to62 public static void main(String args[]) {63 Graph1 g = new Graph1(4);64 65 g.addEdge(0, 1);66 g.addEdge(0, 2);67 g.addEdge(1, 2);68 g.addEdge(2, 0);69 g.addEdge(2, 3);70 g.addEdge(3, 3);71 72 System.out.println("Following is Breadth First Traversal " + "(starting from vertex 2)");73 g.BFS();74 }75 }
4、复杂度分析:
算法借助了一个邻接表和队列,故它的空问复杂度为O(V)。 遍历图的过程实质上是对每个顶点查找其邻接点的过程,其耗费的时间取决于所采用结构。 邻接表表示时,查找所有顶点的邻接点所需时间为O(E),访问顶点的邻接点所花时间为O(V),此时,总的时间复杂度为O(V+E)。
求助一个graph.java主程序及子程序,并请讲一讲它实现的功能
我有一个主程序NOTEBO0K.java
主程序中设置了一个“字体”选择菜单fontmenuitem
我想在按下fontmenuitem菜单时就可弹出字体对话框font
假如字体对话框在子程序FontDialog.java中实现。
在主程序NOTEBO0K.java:
public class NOTEBO0K(){
.....
//设置“字体”选择菜单fontmenuitem的事件
ActionListener fontsetting=new ActionListener(){
public void actionPerformed(ActionEvent e)
{
FontDialog font=new FontDialog(this);//想在这里引入字体对话框
font.setLocation(80,50);
font.setModal(true);
font.show();
}
};
fontmenuitem.addActionListener(fontsetting);
}
在//FontDialog.java
public class FontDialog extends JDialog
{
......
JTextField nameTxt=new JTextField("Default");//nameTxt来表示取得的字体名
JTextField styleTxt=new JTextField("Plain");
JTextField sizeTxt=new JTextField("15");
......... //选择字体的过程
String name=nameTxt.getText();//name来表示取得的字体名
int style=((Integer)styleTbl.get(styleTxt.getText())).intValue();
//style来表示取得的字体类型
String sizeStr=sizeTxt.getText().trim();//sizeStr来表示取得的字体大小
......
}
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