java实现图论中的经典算法

1.最短路的笛杰斯特拉算法/**** @author Administrator*///这个算法用来解决无向图中任意两点的最短路径，同时输出路径(起点到所有点的) public class MinPath {public static String dijkstra(int[][] W1, int start, int end) {System.out.println("起点:" + start + "终点:" + end);boolean[] isLabel = new boolean[W1[0].length];// 是否标号int[] indexs = new int[W1[0].length];// 所有标号的点的下标集合，以标号的先后顺序进行存储，实际上是一个以数组表示的栈int i_count = -1;// 栈的顶点int[] distance = W1[start].clone();// v0到各点的最短距离的初始值int index = start;// 从初始点开始int presentShortest = 0;// 当前临时最短距离indexs[++i_count] = index;// 把已经标号的下标存入下标集中isLabel[index] = true;while (i_count < W1[0].length) {// 第一步：得到与原点最近的某个点int min = Integer.MAX_V ALUE;for (int i = 0; i < distance.length; i++) {if (!isLabel[i] && distance[i] != -1 && i != index) {// 如果到这个点有边,并且没有被标号if (distance[i] < min) {min = distance[i];index = i;// 把下标改为当前下标}}}i_count = i_count + 1;if (i_count == W1[0].length) {break;}isLabel[index] = true;// 对点进行标号indexs[i_count] = index;// 把已经标号的下标存入下标集中if (W1[indexs[i_count - 1]][index] == -1|| presentShortest + W1[indexs[i_count - 1]][index] > distance[index]) { // 如果两个点没有直接相连，或者两个点的路径大于最短路径presentShortest = distance[index];} else {presentShortest += W1[indexs[i_count - 1]][index];}// 第二步：加入vi后，重新计算distance中的距离for (int i = 0; i < distance.length; i++) {// 如果vi到那个点有边，则v0到后面点的距离加if (distance[i] == -1 && W1[index][i] != -1) {// 如果以前不可达，则现在可达了distance[i] = presentShortest + W1[index][i];} else if (W1[index][i] != -1 && presentShortest + W1[index][i] < distance[i]) {// 如果以前可达，但现在的路径比以前更短，则更换成更短的路径distance[i] = presentShortest + W1[index][i];}}}getRoute(W1, indexs, end);return "最短距离是：" + (distance[end] - distance[start]);}public static void main(String[] args) {// 建立一个权值矩阵int[][] W1 = { // 测试数据1{0, 1, 4, -1, -1, -1},{1, 0, 2, 7, 5, -1},{4, 2, 0, -1, 1, -1},{-1, 7, -1, 0, 3, 2},{-1, 5, 1, 3, 0, 6},{-1, -1, -1, 2, 6, 0}};// System.out.println("f" + W1[0][4]);int[][] W = { // 测试数据2{0, 1, 3, 4},{1, 0, 2, -1},{3, 2, 0, 5},{4, -1, 5, 0}};System.out.println(dijkstra(W1, 5, 0)); // (int[][] W1, int start, int end) }// indexs:1,0,2,4,3,5 放顶点的顺序// end:最后要的顶点名称:5// routeLength:长度:8/*** seven 输出路径(起点到所有点的)*/public static String getRoute(int[][] WW, int[] indexs, int end) {String[] routeArray = new String[indexs.length];for (int i = 0; i < routeArray.length; i++) {routeArray[i] = "";}//自己的路线routeArray[indexs[0]] = indexs[0] + "";for (int i = 1; i < indexs.length; i++) {//看该点与前面所有点的连接线中的最短路径，然后得到该最短路径到底是连接了哪个点，进而此点的route就是找出那点的route+此点int[] thePointDis = WW[indexs[i]];int prePoint = 0;int tmp = 9999;for (int j = 0; j < thePointDis.length; j++) {boolean chooseFlag = false;//边的距离最短，而且，所连的点在前面的点当中for (int m = 0; m < i; m++) {if (j == indexs[m]) {chooseFlag = true;}}if (chooseFlag == false) {continue;}if (thePointDis[j] < tmp && thePointDis[j] > 0) {prePoint = j;tmp = thePointDis[j];}}routeArray[indexs[i]] = routeArray[prePoint] + indexs[i];}for (int i = 0; i < routeArray.length; i++) {System.out.println(routeArray[i]);}return "";}}2.最小生成树的Kruskal算法/**** @author Administrator*/public class MinTree {/*** @param args*/private int[][] List_Graph;//图的邻接矩阵private int[] Label;private int[] Weight;private int[] Index_1;private int[] Index_2;private String Result;public Class_Kruskal()//使用构造函数进行初始化数据{List_Graph = new int[][]{{0, 1, 4, 5},{1, 0, 32768, 2},{4, 32768, 0, 3},{5, 2, 3, 0}};//初始化graphic中点和点的距离，32767表示距离无穷大//另外一个测试用例是/*{0,1,2,32767,32767},{1,0,2,4,3},{2,2,0,4,4},{32767,4,4,0,2},{32767,3,4,2,0}*/Label = new int[List_Graph.length];for (int i = 0; i < Label.length; i++)//初始化标记{Label[i] = i;}int j = (int) (Label.length + 1) * Label.length / 2;//这里应该是-1吧,完全图的边数Weight = new int[j];//用于存储待排序边的权值，数组长度m＝（n+1）*n*0.5,其中节点个数为nIndex_1 = new int[j];//用于存储边的两个节点Index_2 = new int[j];Result = "最小生成树的边是：" + "\n";//记录最小生成树的边}public String Get_Result()//获得变量Result{return Result;}//把边按权排序,graphic是List_Graphicpublic int[] sort() {int[] a;int index = 0;for (int i = 0; i < Label.length; i++) {for (int j = i + 1; j < Label.length; j++) {if (List_Graph[i][j] < 32767) {Weight[index] = List_Graph[i][j];Index_1[index] = i;Index_2[index] = j;index = index + 1;}}}a = new int[index - 1];a = Address_Sort(Weight, Weight.length);return a;}public int[] Address_Sort(int[] a, int n)//地址排序{int[] Res = new int[n];for (int i = 0; i < n; i++) {Res[i] = i;}int t;int k;for (int j = 0; j < n - 1; j++) {for (int i = 0; i < n - j - 1; i++) {if (a[i] >= a[i + 1]) {//冒泡法k = a[i];a[i] = a[i + 1];a[i + 1] = k;t = Res[i];Res[i] = Res[i + 1];Res[i + 1] = t;}}}return Res;}public void Min_Tree()//求最小生成树{int[] tag = new int[Weight.length];tag = sort();int i = 0;while (!Judge(Label))//Judge函数判断标记是否都是0{if (Label[Index_1[tag[i]]] != Label[Index_2[tag[i]]]) {//两个点不同Result = Result + Index_1[tag[i]] + "---" + Index_2[tag[i]] + "\n";if (Label[Index_1[tag[i]]] < Label[Index_2[tag[i]]]) {for (int k = 0; k < Label.length; k++) {if (Label[k] == Label[Index_2[tag[i]]]) {Label[k] = Label[Index_1[tag[i]]];}}} else {for (int k = 0; k < Label.length; k++) {if (Label[k] == Label[Index_1[tag[i]]]) {Label[k] = Label[Index_2[tag[i]]];}}}} else {i = i + 1;}}}public boolean Judge(int[] a)//判断标记是否都是0{for (int i = 0; i < a.length; i++) {if (a[i] != 0) {return false;}}return true;}public static void main(String[] args) //主函数{Class_Kruskal CK = new Class_Kruskal();CK.Min_Tree();System.out.println(CK.Get_Result());} }。