java实现乘地铁方案的最优选择（票价，距离）

java实现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 subA = new LinkedList();

private static LinkedList subB = new LinkedList();

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 plots = new HashSet();

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 3; 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 {

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 vertex;// 按顺序保存顶点s

private List edges;

public int[][] getSubTrainMatrix() {

return subTrainMatrix;

}

public void setVertex(List vertices) {

this.vertex = vertices;

}

public List getVertex() {

return vertex;

}

public List getEdges() {

return edges;

}

public int getVertexSize() {

return this.vertex.size();

}

public int vertexCount() {

return subTrainMatrix.length;

}

@Override

public 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();

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 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) {CWBNsbIOBa

int i = getPosInvertex(start);

int j = getPosInvertex(stop);

return this.subTrainMatrix[i][j];

} // 返边的权值

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 vertices = new ArrayList();

vertices.add("A");

vertices.add("B");

vertices.add("C");

vertices.add("D");

vertices.add("E");

graph = new SubTrainMap(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 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 vertices = new ArrayList();

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(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。

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