public class CriticalPath {/** 边 */static class Edge{/** 权重 */int weight;/** 出度指向的点 */int toVertex;Edge next;public Edge(int weight, int toVertex, Edge next) {this.weight = weight;this.toVertex = toVertex;this.next = next;}}/** 顶点 */static class Vertex{/** 入度 数量 */int inNumber;/** 顶点信息 */Integer data;/** 第一条边 */Edge firstEdge;public Vertex(int inNumber, Integer data, Edge firstEdge) {this.inNumber = inNumber;this.data = data;this.firstEdge = firstEdge;}}static void criticalPath(List<Vertex> graph){//顶点数量int length = graph.size();//边数量int numOfEdges = 0;for (Vertex vertex : graph) {Edge edge = vertex.firstEdge;while (edge!=null){numOfEdges ++;edge = edge.next;}}//事件最早发生时间int[] etv = new int[length];//事件最晚发生时间int[] ltv = new int[length];//活动最早发生时间int[] ete = new int[numOfEdges];//活动最晚发生时间int[] lte = new int[numOfEdges];//1. 通过拓扑排序求 etv 「事件最早发生时间」//etvStack 用于储存拓扑排序后的顺序Stack<Vertex> etvStack = new Stack<>();//stack 用于拓扑排序Stack<Vertex> stack = new Stack<>();for (Vertex vertex : graph) {if (vertex.inNumber == 0){stack.push(vertex);}}while (!stack.isEmpty()){Vertex pop = stack.pop();//储存拓扑排序后的结构etvStack.push(pop);//遍历出度Edge edge = pop.firstEdge;while (edge != null){Vertex vertex = graph.get(edge.toVertex);vertex.inNumber --;if (vertex.inNumber == 0){stack.push(vertex);}//赋值更大的距离给 etvif (etv[pop.data] + edge.weight > etv[edge.toVertex]){etv[edge.toVertex] = etv[pop.data] + edge.weight;}edge = edge.next;}}//2.通过 etv 反向推导求出 ltv「事件最晚发生时间」System.out.println("====etv====");for (int i = 0; i < etv.length; i++) {System.out.print("V"+i +" = "+etv[i]+" ");}System.out.println();//初始化 ltvInteger endVertex = etvStack.peek().data;for (int i = 0; i < ltv.length; i++) {ltv[i] = etv[endVertex];}while (!etvStack.isEmpty()) {Vertex pop = etvStack.pop();Edge edge = pop.firstEdge;while (edge != null) {//赋值更小的距离给 ltvif (ltv[pop.data] > ltv[edge.toVertex] - edge.weight) {ltv[pop.data] = ltv[edge.toVertex] - edge.weight;}edge = edge.next;}}System.out.println("====ltv====");for (int i = 0; i < ltv.length; i++) {System.out.print("V"+i +" = "+ltv[i]+" ");}System.out.println();//3. 通过 etv 求 eteint index = 0;for (Vertex vertex : graph) {Edge edge = vertex.firstEdge;while (edge != null){ete[index++] = etv[vertex.data];edge = edge.next;}}System.out.println("====ete====");for (int i = 0; i < ete.length; i++) {System.out.print("E"+i +" = "+ete[i]+" ");}System.out.println();//4. 通过 ltv 求 lteindex = 0;for (Vertex vertex : graph) {Edge edge = vertex.firstEdge;while (edge != null){lte[index++] = ltv[edge.toVertex] - edge.weight;edge = edge.next;}}System.out.println("====lte====");for (int i = 0; i < lte.length; i++) {System.out.print("E"+i +" = "+lte[i]+" ");}System.out.println();//5. 用 lte - ete 求关键路径 System.out.println("====关键路径====");for (int i = 0; i < ete.length; i++) {if (lte[i] - ete[i] == 0) {System.out.print("E"+i+" ");}}return ;}/** 测试 */public static void main(String[] args) {char[] vertices = new char[]{'A','B','C','D','E','F','G'};Edge e3 = new Edge(2, 4, null);Edge e2 = new Edge(1, 3, e3);Edge e1 = new Edge(3, 2, e2);Edge e0 = new Edge(2, 1, e1);Edge e4 = new Edge(1, 5, null);Edge e5 = new Edge(1, 5, null);Edge e6 = new Edge(1, 5, null);Edge e7 = new Edge(1, 5, null);Edge e8 = new Edge(2, 6, null);Vertex a = new Vertex(0, 0, e0);Vertex b = new Vertex(1, 1, e4);Vertex c = new Vertex(1, 2, e5);Vertex d = new Vertex(1, 3, e6);Vertex e = new Vertex(1, 4, e7);Vertex f = new Vertex(4, 5, e8);Vertex g = new Vertex(1, 6, null);ArrayList<Vertex> graph = new ArrayList<>();graph.add(a);graph.add(b);graph.add(c);graph.add(d);graph.add(e);graph.add(f);graph.add(g);criticalPath(graph);}
}

====etv====
V0 = 0 V1 = 2 V2 = 3 V3 = 1 V4 = 2 V5 = 4 V6 = 6
====ltv====
V0 = 0 V1 = 3 V2 = 3 V3 = 3 V4 = 3 V5 = 4 V6 = 6
====ete====
E0 = 0 E1 = 0 E2 = 0 E3 = 0 E4 = 2 E5 = 3 E6 = 1 E7 = 2 E8 = 4
====lte====
E0 = 1 E1 = 0 E2 = 2 E3 = 1 E4 = 3 E5 = 3 E6 = 3 E7 = 3 E8 = 4
====关键路径====
E1 E5 E8