乘地铁方案的最优选择（票价，距离）

初始问题描述：

       已知2条地铁线路，其中A为环线，B为东西向线路，线路都是双向的。经过的站点名分别如下，两条线交叉的换乘点用T1、T2表示。编写程序，任意输入两个站点名称，输出乘坐地铁最少需要经过的车站数量（含输入的起点和终点，换乘站点只计算一次）。
地铁线A（环线）经过车站：A1 A2 A3 A4 A5 A6 A7 A8 A9 T1 A10 A11 A12 A13 T2 A14 A15 A16 A17 A18
地铁线B（直线）经过车站：B1 B2 B3 B4 B5 T1 B6 B7 B8 B9 B10 T2 B11 B12 B13 B14 B15

该特定条件下的实现：

package com.patrick.bishi;import java.util.HashSet;import java.util.LinkedList;import java.util.Scanner;import java.util.Set;/** * 获取两条地铁线上两点间的最短站点数 *  * @author patrick *  */public class SubTrain {private static LinkedList<String> subA = new LinkedList<String>();private static LinkedList<String> subB = new LinkedList<String>();public static void main(String[] args) {String sa[] = { "A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9","T1", "A10", "A11", "A12", "A13", "T2", "A14", "A15", "A16","A17", "A18" };String sb[] = { "B1", "B2", "B3", "B4", "B5", "T1", "B6", "B7", "B8","B9", "B10", "T2", "B11", "B12", "B13", "B14", "B15" };Set<String> plots = new HashSet<String>();for (String t : sa) {plots.add(t);subA.add(t);}for (String t : sb) {plots.add(t);subB.add(t);}Scanner in = new Scanner(System.in);String input = in.nextLine();String trail[] = input.split("\\s");String src = trail[0];String dst = trail[1];if (!plots.contains(src) || !plots.contains(dst)) {System.err.println("no these plot!");return;}int len = getDistance(src, dst);System.out.printf("The shortest distance between %s and %s is %d", src,dst, len);}// 经过两个换乘站点后的距离public static int getDist(String src, String dst) {int len = 0;int at1t2 = getDistOne("T1", "T2");int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1") + 1;int a = 0;if (src.equals("T1")) {a = getDistOne(dst, "T2");len = a + bt1t2 - 1;// two part must more 1} else if (src.equals("T2")) {a = getDistOne(dst, "T1");len = a + bt1t2 - 1;} else if (dst.equals("T1")) {a = getDistOne(src, "T2");len = a + at1t2 - 1;} else if (dst.equals("T2")) {a = getDistOne(src, "T1");len = a + at1t2 - 1;}return len;}// 获得一个链表上的两个元素的最短距离private static int getDistOne(String src, String dst) {int aPre, aBack, aLen, len, aPos, bPos;aPre = aBack = aLen = len = 0;aLen = subA.size();if ("T1".equals(src) && "T2".equals(dst)) {int a = subA.indexOf("T1");int b = subA.indexOf("T2");int at1t2 = (b - a) > (a + aLen - b) ? (a + aLen - b) : (b - a);int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1");len = at1t2 > bt1t2 ? bt1t2 : at1t2;} else if (subA.contains(src) && subA.contains(dst)) {aPos = subA.indexOf(src);bPos = subA.indexOf(dst);if (aPos > bPos) {aBack = aPos - bPos;aPre = aLen - aPos + bPos;len = aBack > aPre ? aPre : aBack;} else {aPre = bPos - aPos;aBack = aLen - bPos + aPos;len = aBack > aPre ? aPre : aBack;}} else if (subB.contains(src) && subB.contains(dst)) {aPos = subB.indexOf(src);bPos = subB.indexOf(dst);len = aPos > bPos ? (aPos - bPos) : (bPos - aPos);} else {System.err.println("Wrong!");}return len + 1;}public static int getDistance(String src, String dst) {int aPre, aBack, len, aLen;aPre = aBack = len = aLen = 0;aLen = subA.size();int a = subA.indexOf("T1");int b = subA.indexOf("T2");int at1t2 = (b - a) > (a + aLen - b) ? (a + aLen - b) : (b - a);int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1");if ((subA.contains(src) && subA.contains(dst))|| (subB.contains(src) && subB.contains(dst))) {len = getDistOne(src, dst);if (src.equals("T1") || src.equals("T2") || dst.equals("T1")|| dst.equals("T2")) {int t = getDist(src, dst);len = len > t ? t : len;}} else {int at1 = getDist(src, "T1");int at2 = getDist(src, "T2");int bt1 = getDist(dst, "T1");int bt2 = getDist(dst, "T2");aPre = at1 + bt1 - 1;aBack = at2 + bt2 - 1;len = aBack > aPre ? aPre : aBack;aPre = at1t2 + at1 + bt2 - 2;aBack = bt1t2 + at2 + bt1 - 2;int tmp = aBack > aPre ? aPre : aBack;len = len > tmp ? tmp : len;}return len;}}

通用乘地铁方案的实现（最短距离利用Dijkstra算法）：

package com.patrick.bishi;import java.util.ArrayList;import java.util.List;import java.util.Scanner;/** * 地铁中任意两点的最有路径 *  * @author patrick *  */public class SubTrainMap<T> {protected int[][] subTrainMatrix; // 图的邻接矩阵，用二维数组表示private static final int MAX_WEIGHT = 99; // 设置最大权值，设置成常量private int[] dist;private List<T> vertex;// 按顺序保存顶点sprivate List<Edge> edges;public int[][] getSubTrainMatrix() {return subTrainMatrix;}public void setVertex(List<T> vertices) {this.vertex = vertices;}public List<T> getVertex() {return vertex;}public List<Edge> getEdges() {return edges;}public int getVertexSize() {return this.vertex.size();}public int vertexCount() {return subTrainMatrix.length;}@Overridepublic String toString() {String str = "邻接矩阵：\n";int n = subTrainMatrix.length;for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++)str += this.subTrainMatrix[i][j] == MAX_WEIGHT ? " $" : " "+ this.subTrainMatrix[i][j];str += "\n";}return str;}public SubTrainMap(int size) {this.vertex = new ArrayList<T>();this.subTrainMatrix = new int[size][size];this.dist = new int[size];for (int i = 0; i < size; i++) { // 初始化邻接矩阵for (int j = 0; j < size; j++) {this.subTrainMatrix[i][j] = (i == j) ? 0 : MAX_WEIGHT;// 无向图}}}public SubTrainMap(List<T> vertices) {this.vertex = vertices;int size = getVertexSize();this.subTrainMatrix = new int[size][size];this.dist = new int[size];for (int i = 0; i < size; i++) { // 初始化邻接矩阵for (int j = 0; j < size; j++) {this.subTrainMatrix[i][j] = (i == j) ? 0 : MAX_WEIGHT;}}}/** * 获得顶点在数组中的位置 *  * @param s * @return */public int getPosInvertex(T s) {return vertex.indexOf(s);}public int getWeight(T start, T stop) {int i = getPosInvertex(start);int j = getPosInvertex(stop);return this.subTrainMatrix[i][j];} // 返<vi,vj>边的权值public void insertEdge(T start, T stop, int weight) { // 插入一条边int n = subTrainMatrix.length;int i = getPosInvertex(start);int j = getPosInvertex(stop);if (i >= 0 && i < n && j >= 0 && j < n&& this.subTrainMatrix[i][j] == MAX_WEIGHT && i != j) {this.subTrainMatrix[i][j] = weight;this.subTrainMatrix[j][i] = weight;}}public void addEdge(T start, T dest, int weight) {this.insertEdge(start, dest, weight);}public void removeEdge(String start, String stop) { // 删除一条边int i = vertex.indexOf(start);int j = vertex.indexOf(stop);if (i >= 0 && i < vertexCount() && j >= 0 && j < vertexCount()&& i != j)this.subTrainMatrix[i][j] = MAX_WEIGHT;}@SuppressWarnings("unused")private static void newGraph() {List<String> vertices = new ArrayList<String>();vertices.add("A");vertices.add("B");vertices.add("C");vertices.add("D");vertices.add("E");graph = new SubTrainMap<String>(vertices);graph.addEdge("A", "B", 5);graph.addEdge("A", "D", 2);graph.addEdge("B", "C", 7);graph.addEdge("B", "D", 6);graph.addEdge("C", "D", 8);graph.addEdge("C", "E", 3);graph.addEdge("D", "E", 9);}private static SubTrainMap<String> graph;/** 打印顶点之间的距离 */public void printL(int[][] a) {for (int i = 0; i < a.length; i++) {for (int j = 0; j < a.length; j++) {System.out.printf("%4d", a[i][j]);}System.out.println();}}public static void main(String[] args) {// newGraph();String sa[] = { "A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9","T1", "A10", "A11", "A12", "A13", "T2", "A14", "A15", "A16","A17", "A18" };String sb[] = { "B1", "B2", "B3", "B4", "B5", "T1", "B6", "B7", "B8","B9", "B10", "T2", "B11", "B12", "B13", "B14", "B15" };List<String> vertices = new ArrayList<String>();for (String t : sa) {if (!vertices.contains(t)) {vertices.add(t);}}for (String t : sb) {if (!vertices.contains(t)) {vertices.add(t);}}graph = new SubTrainMap<String>(vertices);for (int i = 0; i < sa.length - 1; i++)graph.addEdge(sa[i], sa[i + 1], 1);graph.addEdge(sa[0], sa[sa.length - 1], 1);for (int i = 0; i < sb.length - 1; i++)graph.addEdge(sb[i], sb[i + 1], 1);Scanner in = new Scanner(System.in);System.out.println("请输入起始站点：");String start = in.nextLine().trim();System.out.println("请输入目标站点：");String stop = in.nextLine().trim();if (!graph.vertex.contains(start) || !graph.vertex.contains(stop)) {System.out.println("地图中不包含该站点！");return;}int len = graph.find(start, stop) + 1;// 包含自身站点System.out.println(start + " -> " + stop + " 经过的站点数为： " + len);}public int find(T start, T stop) {int startPos = getPosInvertex(start);int stopPos = getPosInvertex(stop);if (startPos < 0 || startPos > getVertexSize())return MAX_WEIGHT;String[] path = dijkstra(startPos);System.out.println("从" + start + "出发到" + stop + "的最短路径为："+ path[stopPos]);return dist[stopPos];}// 单元最短路径问题的Dijkstra算法private String[] dijkstra(int vertex) {int n = dist.length - 1;String[] path = new String[n + 1]; // 存放从start到其他各点的最短路径的字符串表示for (int i = 0; i <= n; i++)path[i] = new String(this.vertex.get(vertex) + "-->"+ this.vertex.get(i));boolean[] visited = new boolean[n + 1];// 初始化for (int i = 0; i <= n; i++) {dist[i] = subTrainMatrix[vertex][i];// 到各个顶点的距离，根据顶点v的数组初始化visited[i] = false;// 初始化访问过的节点，当然都没有访问过}dist[vertex] = 0;visited[vertex] = true;for (int i = 1; i <= n; i++) {// 将所有的节点都访问到int temp = MAX_WEIGHT;int visiting = vertex;for (int j = 0; j <= n; j++) {if ((!visited[j]) && (dist[j] < temp)) {temp = dist[j];visiting = j;}}visited[visiting] = true; // 将距离最近的节点加入已访问列表中for (int j = 0; j <= n; j++) {// 重新计算其他节点到指定顶点的距离if (visited[j]) {continue;}int newdist = dist[visiting] + subTrainMatrix[visiting][j];// 新路径长度，经过visiting节点的路径if (newdist < dist[j]) {// dist[j] 变短dist[j] = newdist;path[j] = path[visiting] + "-->" + this.vertex.get(j);}}// update all new distance}// visite all nodes// for (int i = 0; i <= n; i++)// System.out.println("从" + vertex + "出发到" + i + "的最短路径为：" + path[i]);// System.out.println("=====================================");return path;}/** * 图的边 *  * @author patrick *  */class Edge {private T start, dest;private int weight;public Edge() {}public Edge(T start, T dest, int weight) {this.start = start;this.dest = dest;this.weight = weight;}public String toString() {return "(" + start + "," + dest + "," + weight + ")";}}}

        图中各边的权可以是距离也可以是票价，初始化的方案决定实现的目标。最短路径计算也可以用Floyd算法实现。欢迎其他人讨论和提供实现。