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邻接矩阵存储的图
<span style="font-size:24px;">java实现用邻接矩阵(相邻矩阵)实现图,缺点是矩阵中大量的0元素会耗费大量的存储空间</span> public class Graph { final int MAX_VERTEX = 10;// 最多10个顶点 Vertex[] vertex;// 顶点数组 int[][] adjacency;// 邻接矩阵 int numOfVertex;// 当前图中顶点的数量 public Graph() {// 构造器 vertex = new Vertex[MAX_VERTEX]; adjacency = new int[MAX_VERTEX][MAX_VERTEX]; numOfVertex = 0; // 将邻接矩阵初始化 for (int i = 0; i < MAX_VERTEX; i++) { for (int j = 0; j < MAX_VERTEX; j++) adjacency[i][j] = 0; } } // 添加顶点 public void addVertex(char v) { vertex[numOfVertex++] = new Vertex(v); } //无向图 添加边 public void addEdge(int start, int end) { adjacency[start][end] = 1; adjacency[end][start] = 1; } //有向图添加边 public void addEdge1(int start,int end){ adjacency[start][end] = 1; } // 打印某个顶点 public void showVertex(int index) { System.err.print(vertex[index].label); } // 打印邻接矩阵 public void show() { for (int i = 0; i < MAX_VERTEX; i++) { for (int j = 0; j < MAX_VERTEX; j++) { if (j == MAX_VERTEX - 1) System.out.println(adjacency[i][j] + " "); else System.out.print(adjacency[i][j] + " "); } } } /** * 找到与某一顶点邻接而未被访问的顶点,如何做? * 在邻接矩阵中,找到指定顶点所在的行,从第一列开始向后寻找值为1的列,列号是邻接顶点的号码,检查此顶点是否访问过。 * 如果该行没有值为1而又未访问过的点,则此顶点的邻接点都访问过了。 */ public int getUnVisitedVertex(int index) { for (int i = 0; i < numOfVertex; i++) if (adjacency[index][i] == 1 && vertex[i].wasVisited == false) return i; return -1; } // 图的深度优先遍历 public void dfs() { vertex[0].wasVisited = true;// 从头开始访问 showVertex(0); Stack stack = new Stack(); stack.push(0); /** * 1.用peek()方法获取栈顶的顶点 2.试图找到这个顶点的未访问过的邻接点 3.如果没有找到这样的顶点,出栈 4.如果找到,访问之,入栈 */ while (!stack.isEmpty()) { int index = getUnVisitedVertex(stack.peek()); if (index == -1)// 没有这个顶点 stack.pop(); else { vertex[index].wasVisited = true; showVertex(index); stack.push(index); } } // 栈为空,遍历结束,标记位重新初始化 for (int i = 0; i < numOfVertex; i++) vertex[i].wasVisited = false; } // 图的广度优先遍历 public void bfs() { vertex[0].wasVisited = true; showVertex(0); Queue queue = new Queue(); queue.insert(0); int v2; while (!queue.isEmpty()) {// 直到队列为空 int v1 = queue.remove(); // 直到点v1没有未访问过的邻接点 while ((v2 = getUnVisitedVertex(v1)) != -1) { // 取到未访问过的点,访问之 vertex[v2].wasVisited = true; showVertex(v2); queue.insert(v2); } } for (int i = 0; i < numOfVertex; i++) vertex[i].wasVisited = false; } // 最小生成树 minimum spanning tree public void mst() { vertex[0].wasVisited = true; Stack stack = new Stack(); stack.push(0); while (!stack.isEmpty()) { int currentVertex = stack.peek(); int v = getUnVisitedVertex(currentVertex); if (v == -1) stack.pop(); else { vertex[v].wasVisited = true; stack.push(v); //当前顶点与下一个未访问过的邻接点 showVertex(currentVertex); showVertex(v); System.out.print(" "); } } for (int i = 0; i < numOfVertex; i++) vertex[i].wasVisited = false; } public static void main(String[] args) { Graph graph = new Graph(); graph.addVertex('A'); graph.addVertex('B'); graph.addVertex('C'); graph.addVertex('D'); graph.addEdge(1, 2); graph.addEdge(0, 1); graph.addEdge(2, 3); graph.show(); // graph.dfs(); // graph.bfs(); //graph.mst(); } } // 顶点 class Vertex { char label;// 如A,B,C boolean wasVisited;// 标识是否访问过此顶点 public Vertex(char vertex) { this.label = vertex; wasVisited = false; } } // 栈 class Stack { final int MAX_SIZE = 10; int stack[]; int top; public Stack() { stack = new int[MAX_SIZE]; top = -1; } public void push(int idata) { stack[++top] = idata; } public int pop() { return stack[top--]; } public int peek() { return stack[top]; } public boolean isEmpty() { return top == -1; } } // 队列 class Queue { final int SIZE = 10; int[] qarray; int front; int rear; public Queue() { qarray = new int[SIZE]; front = 0; rear = -1; } // 在队尾追加 public void insert(int key) { if (rear == SIZE - 1) rear = -1; qarray[++rear] = key; } // 队头删除数据 public int remove() { int temp = qarray[front++]; if (front == SIZE) front = 0; return temp; } public boolean isEmpty() { return rear + 1 == front || front + SIZE - 1 == rear; } }
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