有向图和拓扑排序Java实现

2022-12-31,,

package practice;

import java.util.ArrayDeque;
import java.util.Iterator;
import java.util.Stack; public class TestMain {
public static void main(String[] args) {
Digraph a = new Digraph(13);
a.addEdge(0, 1);a.addEdge(0, 5);/*a.addEdge(2, 3);*/a.addEdge(2, 0);a.addEdge(3, 2);
/*a.addEdge(3, 5);*/a.addEdge(4, 3);/*a.addEdge(4, 2);*/a.addEdge(5, 4);a.addEdge(6, 0);
a.addEdge(6, 4);a.addEdge(6, 9);a.addEdge(7, 6);/*a.addEdge(7, 8);*/a.addEdge(8, 7);
a.addEdge(8, 9);a.addEdge(9, 10);a.addEdge(9, 11);a.addEdge(10, 12);a.addEdge(11, 4);
a.addEdge(11, 12);/*a.addEdge(12, 9);*/ System.out.println(a); DirectedDFS dfsa = new DirectedDFS(a);
//第一种方法检查环
System.out.println(dfsa.hasCycle());
System.out.println(); Digraph b = new Digraph(13);
b.addEdge(0, 1);b.addEdge(0, 5);b.addEdge(2, 3);b.addEdge(2, 0);b.addEdge(3, 5);
b.addEdge(5, 4);b.addEdge(6, 4);b.addEdge(6, 9);b.addEdge(7, 6);b.addEdge(8, 7);
b.addEdge(9, 10);b.addEdge(9, 11);b.addEdge(9, 12);b.addEdge(11, 12);b.addEdge(0, 6); DirectedDFS dfsb = new DirectedDFS(b);
//第一种方法检查环
System.out.println(dfsb.hasCycle());
for (Integer integer : dfsb.Topological()) {
System.out.print(integer + " ");
}
}
} /*
* 有向图处理
*/
class DirectedDFS {
private boolean[] marked;
private Digraph G;
private Stack<Integer> cycle;
private int []edgeTo; //轨迹
private boolean[] onStack; //当前搜索轨迹
private Stack<Integer> reversePost; //所有顶点的逆后续排列
/*
* 在图处理类里初始化图
*/
public DirectedDFS(Digraph G) {
marked = new boolean[G.V()];
edgeTo = new int[G.V()];
this.G = G;
DirectedCycle(); //检测环
DepthFirstOrder(); //拓扑排序
} /*
* 深度优先搜索
*/
private void dfs(int s) {
marked[s] = true;
for (int w : G.adj(s)) {
if (!marked[w]) {
edgeTo[w] = s;
dfs(w);
}
}
}
/*
* 广度优先搜索
*/
private void bfs(int s) {
ArrayDeque<Integer> deque = new ArrayDeque<Integer>();
deque.add(s);
marked[s] = true;
int c;
while (!deque.isEmpty()) {
c = deque.poll();
for (int w : G.adj(c)) {
if (!marked[w]) {
marked[w] = true;
edgeTo[w] = c;
deque.add(w);
}
}
}
} /*
* 在G中找到s所有可达的顶点(可以用marked()检测)
*/
public void DFS(int s) {
dfs(s);
}
/*
* 在G中找到source中的所有顶点可到达的所有顶点
*/
public void DFS(Iterable<Integer> sources) {
for (Integer integer : sources) {
dfs(integer);
}
}
/*
* v是可达的吗
*/
public boolean marked(int v) {
return marked[v];
} /*
* 检测s到c的的路径
*/
public void DFDPath(int s, int c) {
for (int i = 0; i < marked.length; i++) {
marked[i] = false;
}
dfs(s);
System.out.print(c + "<-");
while (edgeTo[c] != s) {
c = edgeTo[c];
System.out.print(c + "<-");
}
c = edgeTo[c];
System.out.println(c);
} /*
* 检测s到c的的路径(最短)
*/
public void BFDPath(int s, int c) {
for (int i = 0; i < marked.length; i++) {
marked[i] = false;
}
bfs(s);
System.out.print(c + "<-");
while (edgeTo[c] != s) {
c = edgeTo[c];
System.out.print(c + "<-");
}
c = edgeTo[c];
System.out.println(c);
} /*
* 检测该有向图是否有环
* 要标记当前路径,检查 将要检查到的(并已被标记过的点)是否在当前路径
*/ private void DirectedCycle() {
onStack = new boolean[G.V()];
//先将标记归零
for (int i = 0; i < marked.length; i++) {
marked[i] = false;
}
//挨个搜索节点,标记过的节点就不搜了
for (int i = 0; i < G.V(); i++) {
if (!marked[i]) DirectedCycleDfs(i);
}
} private void DirectedCycleDfs(int s) {
onStack[s] = true; //onStack[v] = true 说明将v加入当前路径
marked[s] = true;
for (int w : G.adj(s)) {
if (this.hasCycle()) return;
else if (!marked[w]) {
edgeTo[w] = s;
DirectedCycleDfs(w);
}
else if (onStack[w]) {
//将这条路径保存到cycle
cycle = new Stack<Integer>();
for (int x = s; x != w; x = edgeTo[x])
cycle.push(x); cycle.push(w);
cycle.push(s);
}
}
onStack[s] = false; //将要切换到另一条路径,将v从当前路径剔除
}
//有环返回true无环返回false
public boolean hasCycle() { return cycle != null;}
//返回找到的环
public Iterable<Integer> cycle() { return cycle;} /*
* 排出顶点的深度优先次序的深度优先搜索
* 拓扑排序 即被指向的一定在指向的后面
*/
public void DepthFirstOrder() {
reversePost = new Stack<Integer>();
for (int i = 0; i < marked.length; i++) {
marked[i] = false;
}
for (int i = 0; i < G.V(); i++) {
if (!marked[i]) dfo(i);
}
}
private void dfo(int s) {
marked[s] = true;
for (int w : G.adj(s)) {
if (!marked[w]) {
dfo(w);
}
}
reversePost.push(s); //可以保证被指向的肯定比指向的先进入栈,所以可以拓扑排序
}
/*
* 拓扑排序
*/
public Iterable<Integer> Topological() {
if (hasCycle()) { return null;} //有环则不能拓扑排序
return reversePost;
}
} /*
* 有向图
*/
class Digraph {
private Bag<Integer>[] digraph;
private int V; //点数
private int E; //边数
public Digraph(int V) {
this.V = V;
digraph = (Bag<Integer>[]) new Bag[V];
for (int i = 0; i < V; i++) {
digraph[i] = new Bag<Integer>();
}
} public int V() { return V;}
public int E() { return E;} public void addEdge(int v, int w) {
digraph[v].add(w);
E++;
} public Iterable<Integer> adj(int V) {
return digraph[V];
} public Digraph reverse() {
Digraph result = new Digraph(V);
for (int i = 0; i < V; i++) {
for (Integer integer : digraph[i]) {
result.addEdge(i, integer);
}
}
return result;
} public String toString() {
String s = V + " vertices, " + E + " edges\n";
for (int v = 0; v < V; v++) {
s += v + ": ";
for (Integer integer : this.adj(v)) {
s += integer + " ";
}
s += "\n";
}
return s;
}
} /*
* 背包
*/
class Bag<T> implements Iterable<T> {
Node first; private class Node {
T value;
Node next;
} public void add(T value) {
Node oldfirst = first;
first = new Node();
first.value = value;
first.next = oldfirst;
} @Override
public Iterator<T> iterator() {
return new BagIterator();
} private class BagIterator implements Iterator<T> {
Node node = first; @Override
public boolean hasNext() {
return node != null;
} @Override
public T next() {
T tempt = node.value;
node = node.next;
return tempt;
}
}
}

上面用到的有向图(Digraph b)

有向图和拓扑排序Java实现的相关教程结束。

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