这篇文章主要介绍如何使用java实现乘地铁方案的最优选择，文中介绍的非常详细，具有一定的参考价值，感兴趣的小伙伴们一定要看完！
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s
private 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;
}
@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<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 + ")";
}
}
}以上是“如何使用java实现乘地铁方案的最优选择”这篇文章的所有内容，感谢各位的阅读！希望分享的内容对大家有帮助，更多相关知识，欢迎关注亿速云行业资讯频道！
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