地铁换乘计费系统(计应192陈莉莉第三组)
psp个人开发流程:
PSP阶段 |
预估时间 |
实际所用时间 |
计划 |
13 |
11 |
|
13 |
11 |
开发 |
91 |
96 |
|
10 |
8 |
|
8 |
10 |
|
12 |
10 |
|
8 |
6 |
|
11 |
15 |
|
19 |
25 |
|
10 |
10 |
|
13 |
12 |
报告 |
16 |
12 |
|
5 |
4 |
|
5 |
3 |
|
6 |
5 |
总共花费时间 |
120 |
107 |
计划:
明确需求会和其他相关因素,计算时间成本
开发:
需求分析:
起点到终点之间经过的车站数量
具体设计:
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
具体代码和代码规范:
1 package com.patrick.bishi; 2 3 import java.util.HashSet; 4 5 import java.util.LinkedList; 6 7 import java.util.Scanner; 8 9 import java.util.Set; 10 11 /** 12 13 * 获取两条地铁线上两点间的最短站点数 14 15 * 16 17 * @author patrick 18 19 * 20 21 */ 22 23 public class SubTrain { 24 private static LinkedList subA = new LinkedList(); 25 26 private static LinkedList subB = new LinkedList(); 27 28 public static void main(String[] args) { 29 String sa[] = { "A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9", 30 31 "T1", "A10", "A11", "A12", "A13", "T2", "A14", "A15", "A16", 32 33 "A17", "A18" }; 34 35 String sb[] = { "B1", "B2", "B3", "B4", "B5", "T1", "B6", "B7", "B8", 36 37 "B9", "B10", "T2", "B11", "B12", "B13", "B14", "B15" }; 38 39 Set plots = new HashSet(); 40 41 for (String t : sa) { 42 plots.add(t); 43 44 subA.add(t); 45 46 } 47 48 for (String t : sb) { 49 plots.add(t); 50 51 subB.add(t); 52 53 } 54 55 Scanner in = new Scanner(System.in); 56 57 String input = in.nextLine(); 58 59 String trail[] = input.split("\\s"); 60 61 String src = trail[0]; 62 63 String dst = trail[1]; 64 65 if (!plots.contains(src) || !plots.contains(dst)) { 66 System.err.println("no these plot!"); 67 68 return; 69 70 } 71 72 int len = getDistance(src, dst); 73 74 System.out.printf("The shortest distance between %s and %s is %d", src, 75 76 dst, len); 77 78 } 79 80 // 经过两个换乘站点后的距离 81 82 public static int getDist(String src, String dst) { 83 int len = 0; 84 85 int at1t2 = getDistOne("T1", "T2"); 86 87 int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1") + 1; 88 89 int a = 0; 90 91 if (src.equals("T1")) { 92 a = getDistOne(dst, "T2"); 93 94 len = a + bt1t2 - 1;// two part must more 1 95 96 } else if (src.equals("T2")) { 97 a = getDistOne(dst, "T1"); 98 99 len = a + bt1t2 - 1; 100 101 } else if (dst.equals("T1")) { 102 a = getDistOne(src, "T2"); 103 104 len = a + at1t2 - 1; 105 106 } else if (dst.equals("T2")) { 107 a = getDistOne(src, "T1"); 108 109 len = a + at1t2 - 1; 110 111 } 112 113 return len; 114 115 } 116 117 // 获得一个链表上的两个元素的最短距离 118 119 private static int getDistOne(String src, String dst) { 120 int aPre, aBack, aLen, len, aPos, bPos; 121 122 aPre = aBack = aLen = len = 0; 123 124 aLen = subA.size(); 125 126 if ("T1".equals(src) && "T2".equals(dst)) { 127 int a = subA.indexOf("T1"); 128 129 int b = subA.indexOf("T2"); 130 131 int at1t2 = (b - a) > (a + aLen - b) ? (a + aLen - b) : (b - a); 132 133 int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1"); 134 135 len = at1t2 > bt1t2 ? bt1t2 : at1t2; 136 137 } else if (subA.contains(src) && subA.contains(dst)) { 138 aPos = subA.indexOf(src); 139 140 bPos = subA.indexOf(dst); 141 142 if (aPos > bPos) { 143 aBack = aPos - bPos; 144 145 aPre = aLen - aPos + bPos; 146 147 len = aBack > aPre ? aPre : aBack; 148 149 } else { 150 aPre = bPos - aPos; 151 152 aBack = aLen - bPos + aPos; 153 154 len = aBack > aPre ? aPre : aBack; 155 156 } 157 158 } else if (subB.contains(src) && subB.contains(dst)) { 159 aPos = subB.indexOf(src); 160 161 bPos = subB.indexOf(dst); 162 163 len = aPos > bPos ? (aPos - bPos) : (bPos - aPos); 164 165 } else { 166 System.err.println("Wrong!"); 167 168 } 169 170 return len + 1; 171 172 } 173 174 public static int getDistance(String src, String dst) { 175 int aPre, aBack, len, aLen; 176 177 aPre = aBack = len = aLen = 0; 178 179 aLen = subA.size(); 180 181 int a = subA.indexOf("T1"); 182 183 int b = subA.indexOf("T2"); 184 185 int at1t2 = (b - a) > (a + aLen - b) ? (a + aLen - b) : (b - a); 186 187 int bt1t2 = subB.indexOf("T2") - subB.indexOf("T1"); 188 189 if ((subA.contains(src) && subA.contains(dst)) 190 191 || (subB.contains(src) && subB.contains(dst))) { 192 len = getDistOne(src, dst); 193 194 if (src.equals("T1") || src.equals("T2") || dst.equals("T1") 195 196 || dst.equals("T2")) { 197 int t = getDist(src, dst); 198 199 len = len > t ? t : len; 200 201 } 202 203 } else { 204 int at1 = getDist(src, "T1"); 205 206 int at2 = getDist(src, "T2"); 207 208 int bt1 = getDist(dst, "T1"); 209 210 int bt2 = getDist(dst, "T2"); 211 212 aPre = at1 + bt1 - 1; 213 214 aBack = at2 + bt2 - 1; 215 216 len = aBack > aPre ? aPre : aBack; 217 218 aPre = at1t2 + at1 + bt2 - 2; 219 220 aBack = bt1t2 + at2 + bt1 - 2; 221 222 int tmp = aBack > aPre ? aPre : aBack; 223 224 len = len > tmp ? tmp : len; 225 226 } 227 228 return len; 229 230 } 231 232 } 233 234 通用乘地铁方案的实现(最短距离利用Dijkstra算法): 235 236 package com.patrick.bishi; 237 238 import java.util.ArrayList; 239 240 import java.util.List; 241 242 import java.util.Scanner; 243 244 /** 245 246 * 地铁中任意两点的最有路径 247 248 * 249 250 * @author patrick 251 252 * 253 254 */ 255 256 public class SubTrainMap { 257 protected int[][] subTrainMatrix; // 图的邻接矩阵,用二维数组表示 258 259 private static final int MAX_WEIGHT = 99; // 设置最大权值,设置成常量 260 261 private int[] dist; 262 263 private List vertex;// 按顺序保存顶点s 264 265 private List edges; 266 267 public int[][] getSubTrainMatrix() { 268 return subTrainMatrix; 269 270 } 271 272 public void setVertex(List vertices) { 273 this.vertex = vertices; 274 275 } 276 277 public List getVertex() { 278 return vertex; 279 280 } 281 282 public List getEdges() { 283 return edges; 284 285 } 286 287 public int getVertexSize() { 288 return this.vertex.size(); 289 290 } 291 292 public int vertexCount() { 293 return subTrainMatrix.length; 294 295 } 296 297 @Override 298 299 public String toString() { 300 String str = "邻接矩阵:\n"; 301 302 int n = subTrainMatrix.length; 303 304 for (int i = 0; i < n; i++) { 305 for (int j = 0; j < n; j++) 306 307 str += this.subTrainMatrix[i][j] == MAX_WEIGHT ? " $" : " " 308 309 + this.subTrainMatrix[i][j]; 310 311 str += "\n"; 312 313 } 314 315 return str; 316 317 } 318 319 public SubTrainMap(int size) { 320 this.vertex = new ArrayList(); 321 322 this.subTrainMatrix = new int[size][size]; 323 324 this.dist = new int[size]; 325 326 for (int i = 0; i < size; i++) { // 初始化邻接矩阵 327 328 for (int j = 0; j < size; j++) { 329 this.subTrainMatrix[i][j] = (i == j) ? 0 : MAX_WEIGHT;// 无向图 330 331 } 332 333 } 334 335 } 336 337 public SubTrainMap(List vertices) { 338 this.vertex = vertices; 339 340 int size = getVertexSize(); 341 342 this.subTrainMatrix = new int[size][size]; 343 344 this.dist = new int[size]; 345 346 for (int i = 0; i < size; i++) { // 初始化邻接矩阵 347 348 for (int j = 0; j < size; j++) { 349 this.subTrainMatrix[i][j] = (i == j) ? 0 : MAX_WEIGHT; 350 351 } 352 353 } 354 355 } 356 357 /** 358 359 * 获得顶点在数组中的位置 360 361 * 362 363 * @param s 364 365 * @return 366 367 */ 368 369 public int getPosInvertex(T s) { 370 return vertex.indexOf(s); 371 372 } 373 374 public int getWeight(T start, T stop) { 375 int i = getPosInvertex(start); 376 377 int j = getPosInvertex(stop); 378 379 return this.subTrainMatrix[i][j]; 380 381 } // 返边的权值 382 383 public void insertEdge(T start, T stop, int weight) { // 插入一条边 384 385 int n = subTrainMatrix.length; 386 387 int i = getPosInvertex(start); 388 389 int j = getPosInvertex(stop); 390 391 if (i >= 0 && i < n && j >= 0 && j < n 392 393 && this.subTrainMatrix[i][j] == MAX_WEIGHT && i != j) { 394 this.subTrainMatrix[i][j] = weight; 395 396 this.subTrainMatrix[j][i] = weight; 397 398 } 399 400 } 401 402 public void addEdge(T start, T dest, int weight) { 403 this.insertEdge(start, dest, weight); 404 405 } 406 407 public void removeEdge(String start, String stop) { // 删除一条边 408 409 int i = vertex.indexOf(start); 410 411 int j = vertex.indexOf(stop); 412 413 if (i >= 0 && i < vertexCount() && j >= 0 && j < vertexCount() 414 415 && i != j) 416 417 this.subTrainMatrix[i][j] = MAX_WEIGHT; 418 419 } 420 421 @SuppressWarnings("unused") 422 423 private static void newGraph() { 424 List vertices = new ArrayList(); 425 426 vertices.add("A"); 427 428 vertices.add("B"); 429 430 vertices.add("C"); 431 432 vertices.add("D"); 433 434 vertices.add("E"); 435 436 graph = new SubTrainMap(vertices); 437 438 graph.addEdge("A", "B", 5); 439 440 graph.addEdge("A", "D", 2); 441 442 graph.addEdge("B", "C", 7); 443 444 graph.addEdge("B", "D", 6); 445 446 graph.addEdge("C", "D", 8); 447 448 graph.addEdge("C", "E", 3); 449 450 graph.addEdge("D", "E", 9); 451 452 } 453 454 private static SubTrainMap graph; 455 456 /** 打印顶点之间的距离 */ 457 458 public void printL(int[][] a) { 459 for (int i = 0; i < a.length; i++) { 460 for (int j = 0; j < a.length; j++) { 461 System.out.printf("%4d", a[i][j]); 462 463 } 464 465 System.out.println(); 466 467 } 468 469 } 470 471 public static void main(String[] args) { 472 // newGraph(); 473 474 String sa[] = { "A1", "A2", "A3", "A4", "A5", "A6", "A7", "A8", "A9", 475 476 "T1", "A10", "A11", "A12", "A13", "T2", "A14", "A15", "A16", 477 478 "A17", "A18" }; 479 480 String sb[] = { "B1", "B2", "B3", "B4", "B5", "T1", "B6", "B7", "B8", 481 482 "B9", "B10", "T2", "B11", "B12", "B13", "B14", "B15" }; 483 484 List vertices = new ArrayList(); 485 486 for (String t : sa) { 487 if (!vertices.contains(t)) { 488 vertices.add(t); 489 490 } 491 492 } 493 494 for (String t : sb) { 495 if (!vertices.contains(t)) { 496 vertices.add(t); 497 498 } 499 500 } 501 502 graph = new SubTrainMap(vertices); 503 504 for (int i = 0; i < sa.length - 1; i++) 505 506 graph.addEdge(sa[i], sa[i + 1], 1); 507 508 graph.addEdge(sa[0], sa[sa.length - 1], 1); 509 510 for (int i = 0; i < sb.length - 1; i++) 511 512 graph.addEdge(sb[i], sb[i + 1], 1); 513 514 Scanner in = new Scanner(System.in); 515 516 System.out.println("请输入起始站点:"); 517 518 String start = in.nextLine().trim(); 519 520 System.out.println("请输入目标站点:"); 521 522 String stop = in.nextLine().trim(); 523 524 if (!graph.vertex.contains(start) || !graph.vertex.contains(stop)) { 525 System.out.println("地图中不包含该站点!"); 526 527 return; 528 529 } 530 531 int len = graph.find(start, stop) + 1;// 包含自身站点 532 533 System.out.println(start + " -> " + stop + " 经过的站点数为: " + len); 534 535 } 536 537 public int find(T start, T stop) { 538 int startPos = getPosInvertex(start); 539 540 int stopPos = getPosInvertex(stop); 541 542 if (startPos < 0 || startPos > getVertexSize()) 543 544 return MAX_WEIGHT; 545 546 String[] path = dijkstra(startPos); 547 548 System.out.println("从" + start + "出发到" + stop + "的最短路径为:" 549 550 + path[stopPos]); 551 552 return dist[stopPos]; 553 554 } 555 556 // 单元最短路径问题的Dijkstra算法 557 558 private String[] dijkstra(int vertex) { 559 int n = dist.length - 1; 560 561 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 562 563 for (int i = 0; i <= n; i++) 564 565 path[i] = new String(this.vertex.get(vertex) + "-->" 566 567 + this.vertex.get(i)); 568 569 boolean[] visited = new boolean[n + 1]; 570 571 // 初始化 572 573 for (int i = 0; i <= n; i++) { 574 dist[i] = subTrainMatrix[vertex][i];// 到各个顶点的距离,根据顶点v的数组初始化 575 576 visited[i] = false;// 初始化访问过的节点,当然都没有访问过 577 578 } 579 580 dist[vertex] = 0; 581 582 visited[vertex] = true; 583 584 for (int i = 1; i <= n; i++) {// 将所有的节点都访问到 585 586 int temp = MAX_WEIGHT; 587 588 int visiting = vertex; 589 590 for (int j = 0; j <= n; j++) { 591 if ((!visited[j]) && (dist[j] < temp)) { 592 temp = dist[j]; 593 594 visiting = j; 595 596 } 597 598 } 599 600 visited[visiting] = true; // 将距离最近的节点加入已访问列表中 601 602 for (int j = 0; j <= n; j++) {// 重新计算其他节点到指定顶点的距离 603 604 if (visited[j]) { 605 continue; 606 607 } 608 609 int newdist = dist[visiting] + subTrainMatrix[visiting][j];// 新路径长度,经过visiting节点的路径 610 611 if (newdist < dist[j]) { 612 // dist[j] 变短 613 614 dist[j] = newdist; 615 616 path[j] = path[visiting] + "-->" + this.vertex.get(j); 617 618 } 619 620 }// update all new distance 621 622 }// visite all nodes 623 624 // for (int i = 0; i <= n; i++) 625 626 // System.out.println("从" + vertex + "出发到" + i + "的最短路径为:" + path[i]); 627 628 // System.out.println("====================================="); 629 630 return path; 631 632 } 633 634 /** 635 636 * 图的边 637 638 * 639 640 * @author patrick 641 642 * 643 644 */ 645 646 class Edge { 647 private T start, dest; 648 649 private int weight; 650 651 public Edge() { 652 } 653 654 public Edge(T start, T dest, int weight) { 655 this.start = start; 656 657 this.dest = dest; 658 659 this.weight = weight; 660 661 } 662 663 public String toString() { 664 return "(" + start + "," + dest + "," + weight + ")"; 665 666 } 667 668 } 669 670 }
代码复审:
编写代码不可能一次就成功,需要使用debug进行检查运行,对出现的错误及时进行修改,对代码进行完善。
总结:
由于自身基础薄弱,知识能力有限,所以本次项目对于自己来说还是有些困难的,所以和他人一起协作才完成了这个项目,这也让我明白了结对协作的重要性,以后定加倍努力学习,希望自己在这条路上能够走的更远