老伯伯要带鱼、狗、猫过河到
对岸.，有一条船，只能坐一个人，
老伯每次只能带一样动物过河，当
老伯不在的时侯狗会咬猫，猫会吃
鱼.，请问怎么顺序过河呢？要求：编
写程序，由程序来推出过河的顺序

这怎么用对象来搞啊？？



谢谢，有想法了，稍后贴代码

陈团辉 发表于 2013-3-21 08:52

这个跟百度的一模一样，就是改了一下名字！靠！

1. import java.util.Iterator;
   
2. import java.util.LinkedList;
   
3. 
   
4. public class AcrossTheRiver {
   
5. // 定义三个String对象
   
6. public static final String rabbitName = "Rabbit";
   
7. public static final String wolfName = "Wolf";
   
8. public static final String cabbageName = "Cabbage";
   
9. 
   
10. // 判断两个货物之间关系是否友好 写的麻烦了一点= =..
    
11. public static boolean isFriendly(Goods goods1, Goods goods2) {
    
12. if (goods1 != null) {
    
13. if (goods1.getGoodsName().trim().equals(rabbitName)) {
    
14. if (goods2 == null) {
    
15. return true;
    
16. } else {
    
17. return false;
    
18. }
    
19. } else if (goods1.getGoodsName().trim().equals(wolfName)) {
    
20. if (goods2 == null || goods2.getGoodsName().trim().equals(cabbageName)) {
    
21. return true;
    
22. } else {
    
23. return false;
    
24. }
    
25. } else if (goods1.getGoodsName().trim().equals(cabbageName)) {
    
26. if (goods2 == null || goods2.getGoodsName().trim().equals(wolfName)) {
    
27. return true;
    
28. } else {
    
29. return false;
    
30. }
    
31. } else {
    
32. return false;
    
33. }
    
34. } else {
    
35. return true;
    
36. }
    
37. }
    
38. 
    
39. // 我就直接写在主方法里了
    
40. public static void main(String[] args) {
    
41. boolean isSuccess = false;
    
42. LinkedList<Goods> beforeCrossing = new LinkedList<Goods>();
    
43. LinkedList<Goods> afterCrossing = new LinkedList<Goods>();
    
44. beforeCrossing.add(new Goods(rabbitName));
    
45. beforeCrossing.add(new Goods(cabbageName));
    
46. beforeCrossing.add(new Goods(wolfName));
    
47. while (!isSuccess) {
    
48. Goods goods1 = beforeCrossing.getFirst();
    
49. System.out.println(goods1.getGoodsName() + " 被取走了");
    
50. beforeCrossing.removeFirst();
    
51. if (beforeCrossing.isEmpty()) {
    
52. afterCrossing.addLast(goods1);
    
53. isSuccess = true;
    
54. System.out.println("全部移动完毕！");
    
55. } else {
    
56. Iterator<Goods> it = beforeCrossing.iterator();
    
57. Goods[] beforeCro = new Goods[2];
    
58. for (int i = 0; it.hasNext(); i++) {
    
59. beforeCro[i] = it.next();
    
60. System.out.println(beforeCro[i].getGoodsName() + " 留了下来");
    
61. }
    
62. if (isFriendly(beforeCro[0], beforeCro[1])) {
    
63. if (afterCrossing.isEmpty()) {
    
64. afterCrossing.addLast(goods1);
    
65. System.out.println(goods1.getGoodsName() + " 被成功的放到了对岸");
    
66. } else {
    
67. Goods goods2 = afterCrossing.getFirst();
    
68. if (isFriendly(goods1, goods2)) {
    
69. afterCrossing.addLast(goods1);
    
70. System.out.println(goods1.getGoodsName() + " 被成功的放到了对岸");
    
71. } else {
    
72. beforeCrossing.addLast(goods2);
    
73. afterCrossing.removeFirst();
    
74. System.out.println(goods1.getGoodsName() + " 与 "
    
75. + goods2.getGoodsName() + "并不和睦 于是把 " + goods2.getGoodsName()
    
76. + "带了回来 并将 " + goods1.getGoodsName() + " 留了下来");
    
77. }
    
78. }
    
79. } else {
    
80. beforeCrossing.addLast(goods1);
    
81. System.out.println("很可惜 留下来的两个东西并不和睦 于是 " + goods1.getGoodsName()
    
82. + " 又被放了回去");
    
83. }
    
84. }
    
85. }
    
86. }
    
87. }
    
88. 
    
89. // 货物类
    
90. class Goods {
    
91. // 货物名称
    
92. private String goodsName;
    
93. 
    
94. // 默认构造方法
    
95. public Goods(String goodsName) {
    
96. this.goodsName = goodsName;
    
97. }
    
98. 
    
99. // 获得货物名称
    
100. public String getGoodsName() {
     
101. return goodsName;
     
102. }
     
103. 
     
104. }

复制代码

来个不用程序的：
1.先把猫带过去，回来；
2.再把狗带过去，把猫带回来；
3.把鱼带过去，回来；
4.把猫带过去。