青蛙过河游戏和C#算法
2 using System.Collections.Generic;
3 using System.ComponentModel;
4 using System.Data;
5 using System.Drawing;
6 using System.Linq;
7 using System.Text;
8 using System.Windows.Forms;
9
10 namespace 青蛙过河
11 {
12 public partial class Form1 : Form
13 {
14 public Form1()
15 {
16 InitializeComponent();
17 }
18
19
20 private void button1_Click(object sender, EventArgs e)
21 {
22 //调用算法类
23 FrogOverRiver FrogOverRiver = new FrogOverRiver();
24 this.textBox1.Text = FrogOverRiver.getFrogJumpOrder();
25 }
26 }
27
28 //青蛙过河类
29 class FrogOverRiver
30 {
31
32 //获取青蛙的跳动顺序
33 public string getFrogJumpOrder()
34 {
35 List<Frog> frogQueue = this.initializeFrogQueue();
36 return this.frogJump(frogQueue, 3);
37 }
3839
40 #region 辅助函数
41
42 //初始化青蛙队列
43 private List<Frog> initializeFrogQueue()
44 {
45 List<Frog> frogQueue = new List<Frog>();
46 frogQueue.Add(new Frog(0, "左1", frogDirection.向右, false));
47 frogQueue.Add(new Frog(1, "左2", frogDirection.向右, true));
48 frogQueue.Add(new Frog(2, "左3", frogDirection.向右, true));
49 frogQueue.Add(new Frog(3));
50 frogQueue.Add(new Frog(4, "右1", frogDirection.向左, true));
51 frogQueue.Add(new Frog(5, "右2", frogDirection.向左, true));
52 frogQueue.Add(new Frog(6, "右3", frogDirection.向左, false));
53 return frogQueue;
54 }
55
56 //当一个青蛙跳动后,形成一个新的队列
57 private List<Frog> editFrogQueue(List<Frog> frogQueue, string frogName, int oldEmptyPositonID, int newEmptyPositonID)
58 {
59 List<Frog> newFrogQueue = new List<Frog>();
60 foreach (Frog frog in frogQueue)
61 {
62 Frog newFrog = new Frog(frog);
63
64 if (newFrog.isEmpty)
65 newFrog.position = newEmptyPositonID;
66
67 if (newFrog.frogName == frogName)
68 {
69 newFrog.position = oldEmptyPositonID;
70 }
71
72 newFrog.canJump = false;
73 if ((newEmptyPositonID - newFrog.position) > 0 && (newEmptyPositonID - newFrog.position) < 3 && newFrog.direction == frogDirection.向右)
74 newFrog.canJump = true;
75
76 if ((newFrog.position - newEmptyPositonID) > 0 && (newFrog.position - newEmptyPositonID) < 3 && newFrog.direction == frogDirection.向左)
77 newFrog.canJump = true;
78
79 newFrogQueue.Add(newFrog);
80 }
81 return newFrogQueue;
82 }
83
84 //是否已经完成位置对换,即前三个青蛙的位置都大于3
85 private bool isComplete(List<Frog> frogQueue)
86 {
87 return (frogQueue[0].position > 3 && frogQueue[1].position > 3 && frogQueue[2].position > 3);
88 }
89
90 //是否还有可以跳动的青蛙,只要有可以跳动的,就没有达到最后的状态，但都不可以跳动了也不一定对换完了，这里只是控制递归
91 private bool canFrogJump(List<Frog> frogQueue)
92 {
93 foreach (Frog frog in frogQueue)
94 {
95 if (frog.canJump)
96 return true;
97 }
98 return false;
99 }
100 #endregion
101
102 /// <summary>
103 /// 获取青蛙的跳动步骤
104 /// </summary>
105 /// <param name="frogQueue">当前青蛙队列</param>
106 /// <param name="emptyPositionId">空位置编号</param>
107 /// <returns></returns>
108 public string frogJump(List<Frog> frogQueue, int emptyPositionId)
109 {
110 string frogJumpInfo = "";
111
112 foreach (Frog frog in frogQueue)
113 {
114 //是空位置
115 if (frog.isEmpty)
116 continue;
117 if (!frog.canJump)
118 continue;
119
120 frogJumpInfo = "青蛙" + frog.frogName + " " + frog.direction + "跳到" + (emptyPositionId + 1).ToString() + "\r\n";
121
122 int newPositionId = frog.position;
123 List<Frog> newFrogQueue = this.editFrogQueue(frogQueue, frog.frogName, emptyPositionId, newPositionId);
124
125 //只要能继续跳就递归
126 if (this.canFrogJump(newFrogQueue))
127 {
128 frogJumpInfo += this.frogJump(newFrogQueue, newPositionId);
129 }
130 else
131 {
132 if (this.isComplete(newFrogQueue))
133 {
134 frogJumpInfo = frogJumpInfo + "成功";
135 break;
136 }
137 }
138
139 if (frogJumpInfo.Contains("成功"))
140 break;
141 }
142 //循环结束
143 return frogJumpInfo;
144 }
145
146 }
147
148 #region 辅助类
149 enum frogDirection { 向左, 向右 };
150 //青蛙类
151 class Frog
152 {
153 public string frogName;//青蛙名称
154 public int position; //青蛙位置
155 public frogDirection direction; //青蛙跳动的方向
156 public bool canJump;//是否可以跳
157 public bool isEmpty = false; //是否是空格
158
159 public Frog(int positon, string frogName, frogDirection direction, bool canJump)
160 {
161 this.position = positon;
162 this.frogName = frogName;
163 this.direction = direction;
164 this.canJump = canJump;
165 }
166
167 public Frog(int positon)
168 {
169 this.frogName = "空";
170 this.position = positon;
171 this.canJump = false;
172 this.isEmpty = true;
173 }
174 public Frog(Frog frog)
175 {
176 this.position = frog.position;
177 this.frogName = frog.frogName;
178 this.direction = frog.direction;
179 this.canJump = frog.canJump;
180 this.isEmpty = frog.isEmpty;
181 }
182 }
183 #endregion
184 }
185
186
游戏和代码下载
网上找到的一个算法
private void button1_Click(object sender, EventArgs e)
{
int[] begintArray = new int[] { 1, 1, 1, 0, 2, 2, 2 };
int[] endArray = new int[] { 2, 2, 2, 0, 1, 1, 1 };
//最后的解，最先移动的排在最后
List<string> jumpStep = new List<string>();
jump(begintArray, endArray, jumpStep);
//反转一下得到正序的解
jumpStep.Reverse();
foreach (string step in jumpStep)
Console.WriteLine(step);
}
private bool jump(int[] begintArray, int[] endArray, List<string> jumpStep)
{
if (compareArray(begintArray, endArray))
return true;
int zeroIndex = Array.IndexOf(begintArray, 0);
int[] moveTo = zeroCanMoveTo(begintArray, zeroIndex);
//穷举所有可以移动的位置
foreach (int moveStep in moveTo)
{
int[] newBegin = (int[])begintArray.Clone();
newBegin[zeroIndex] = newBegin[moveStep];
newBegin[moveStep] = 0;
if (jump(newBegin, endArray, jumpStep))
{
string steplog = "";
for (int i = 0; i < newBegin.Length; i++)
steplog += newBegin[i].ToString();
jumpStep.Add(steplog);
return true;
}
}
return false;
}
//判断0可以移动到那些格
private int[] zeroCanMoveTo(int[] begintArray, int zeroIndex)
{
List<int> moveTo = new List<int>();
if (zeroIndex >= 1 && begintArray[zeroIndex - 1] == 1)
moveTo.Add(zeroIndex - 1);
if (zeroIndex >= 2 && begintArray[zeroIndex - 2] == 1)
moveTo.Add(zeroIndex - 2);
if (zeroIndex < begintArray.Length - 1 && begintArray[zeroIndex + 1] == 2)
moveTo.Add(zeroIndex + 1);
if (zeroIndex < begintArray.Length - 2 && begintArray[zeroIndex + 2] == 2)
moveTo.Add(zeroIndex + 2);
return moveTo.ToArray();
}
//比较两个数组是否相等
private bool compareArray(int[] begintArray, int[] endArray)
{
for (int i = 0; i < begintArray.Length; i++)
{
if (begintArray[i] != endArray[i])
return false;
}
return true;
}