題目: 模擬一個微型生態系統。有青草、兔子、狐狸三種生物。可以用網格表示整個環境，每個格子內用不同字母表示不同事物：空地E青草G兔子R狐狸F。系統規則如下：

1）初始狀態隨機出現GRF；

2）若格子為G，下一輪其周邊八個格子中為E的格子中會至少有一個變為G；

3）若格子為R，若其周圍八個格子都沒有G或其中一個出現F，則下一輪此格子為E；否則其中一個為G的格子變為R；

4）若格子為F，其周圍八個格子都沒有R，則下一輪此格子為E；否則其中一個為R的格子變成F；

5）若格子為E，則下一輪會以一定的概率變為G。

請分析並畫出類結構圖，利用面向物件程式設計思想，完成模擬過程。（必須用到類的繼承、虛擬函式）

問題分析：

感覺沒有很大的必要對每一種生物建立一個類，並進行派生，所以這裡我只建立了兩個類，一個世界（world）作為基類，一個生態系統（ShengTai）作為派生類，繼承於基類（world），輸出函式（Display）作為虛擬函式，建立基類指標呼叫派生類的虛擬函式。類結構圖如下：

程式碼實現：（執行環境為Microsoft Visual Studio 2010）

/*初始化生態系統大小為10*10（可以改變），空地變為青草地的概率為60%（可以改變），規則中某些地方指出某些情況下，至少一個改變或者其中一個地方改變，這裡均按照從上到下從左到右的順序，以出現滿足條件的第一個進行改變，其餘不變。*/

#include<iostream>
#include<ctime>
#include<cstdlib>
using namespace std;
const int col = 10;
const int row = 10;//初始化生態系統大小<col*row>
const int TurnGrass = 60;//空地變為青草地的概率
class world
{
public:
	void Init();
	virtual void Display();
};
void world::Init()
{
	cout<<"world init"<<endl;
}
void world::Display()
{
	cout<<"nothing found,world empty!"<<endl;
}
class ShengTai:public world
{
public:
	void Init();//初始化
	void Develop();//演化
	void Display();//輸出狀態
	void Traverse();//遍歷
private:
	int s[col][row];//存放0，1，2，3，分別代表空地青草兔子狐狸
	char life;//存放生物或者空地
};
void ShengTai::Init()
{
	int i,j;
	srand((unsigned)time(NULL));
	for(i=0;i<col;i++)
		for(j=0;j<row;j++)
			s[i][j] = rand()%(3-0+1)+0;
	Traverse();
}
void ShengTai::Traverse()
{
	int i,j;
	for(i=0;i<col;i++)
		for(j=0;j<row;j++)
			switch(s[i][j])
		{
			case 0:life = 'E';break;
			case 1:life = 'G';break;
			case 2:life = 'R';break;
			case 3:life = 'F';break;
			default:cout<<"error traverse"<<endl;
		}
}
void ShengTai::Display()
{
	int i,j;
	for(i=0;i<col;i++)
	{
		for(j=0;j<row;j++)
			switch(s[i][j])
		{
			case 0:cout<<"E"<<" ";break;
			case 1:cout<<"G"<<" ";break;
			case 2:cout<<"R"<<" ";break;
			case 3:cout<<"F"<<" ";break;
			default:cout<<"error display"<<endl;
		}
		cout<<endl;
	}
	cout<<endl;
}
void ShengTai::Develop()
{
	int i,j,turned = 0;
	int tmp[col][row];
	for(i=0;i<col;i++)
		for(j=0;j<row;j++)
			tmp[i][j] = 5;
	for(i=0;i<col;i++)
		for(j=0;j<row;j++)
			switch(s[i][j])
		{
			case 0:
				{
					int k;
					k = rand()%(100-1+1)+1;
					if(k >= TurnGrass)
						tmp[i][j] = 1;
					break;
				}
			case 1:
				{
					turned = 0;
					if(s[i-1][j-1] == 0)
					{
						tmp[i-1][j-1] = 1;
						turned = 1;
					}
					if(s[i-1][j] == 0&&turned == 0)
					{
						tmp[i-1][j] = 1;
						turned = 1;
					}
					if(s[i-1][j+1] == 0&&turned == 0)
					{
						tmp[i-1][j+1] = 1;
						turned = 1;
					}
					if(s[i][j-1] == 0&&turned == 0)
					{
						tmp[i][j-1] = 1;
						turned = 1;
					}
					if(s[i][j+1] == 0&&turned == 0)
					{
						tmp[i][j+1] = 1;
						turned = 1;
					}
					if(s[i+1][j-1] == 0&&turned == 0)
					{
						tmp[i+1][j-1] = 1;
						turned = 1;
					}
					if(s[i+1][j] == 0&&turned == 0)
					{
						tmp[i+1][j] = 1;
						turned = 1;
					}
					if(s[i+1][j+1] == 0&&turned == 0)
					{
						tmp[i+1][j+1] = 1;
						turned = 1;
					}
					break;
				}
			case 2:
				{
					turned = 0;
					if(s[i-1][j-1] == 3)
					{
						tmp[i-1][j-1] = 3;
						turned = 1;
					}
					if(s[i-1][j] == 3&&turned == 0)
					{
						tmp[i-1][j] = 3;
						turned = 1;
					}
					if(s[i-1][j+1] == 3&&turned == 0)
					{
						tmp[i-1][j+1] = 3;
						turned = 1;
					}
					if(s[i][j-1] == 3&&turned == 0)
					{
						tmp[i][j-1] = 3;
						turned = 1;
					}
					if(s[i][j+1] == 3&&turned == 0)
					{
						tmp[i][j+1] = 3;
						turned = 1;
					}
					if(s[i+1][j-1] == 3&&turned == 0)
					{
						tmp[i+1][j-1] = 3;
						turned = 1;
					}
					if(s[i+1][j] == 3&&turned == 0)
					{
						tmp[i+1][j] = 3;
						turned = 1;
					}
					if(s[i+1][j+1] == 3&&turned == 0)
					{
						tmp[i+1][j+1] = 3;
						turned = 1;
					}
					if(turned == 0&&
						s[i-1][j-1] != 1&&s[i-1][j] != 1&&
						s[i-1][j+1] != 1&&s[i][j-1] != 1&&
						s[i][j+1] != 1&&s[i+1][j-1] != 1&&
						s[i+1][j] != 1&&s[i+1][j+1] != 1)
					{
						if(s[i-1][j-1] == 1)
						{
							tmp[i-1][j-1] = 2;
							turned = 1;
						}
						if(s[i-1][j] == 1&&turned == 0)
						{
							tmp[i-1][j] = 2;
							turned = 1;
						}
						if(s[i-1][j+1] == 1&&turned == 0)
						{
							tmp[i-1][j+1] = 2;
							turned = 1;
						}
						if(s[i][j-1] == 1&&turned == 0)
						{
							tmp[i][j-1] = 2;
							turned = 1;
						}
						if(s[i][j+1] == 1&&turned == 0)
						{
							tmp[i][j+1] = 2;
							turned = 1;
						}
						if(s[i+1][j-1] == 1&&turned == 0)
						{
							tmp[i+1][j-1] = 2;
							turned = 1;
						}
						if(s[i+1][j] == 1&&turned == 0)
						{
							tmp[i+1][j] = 2;
							turned = 1;
						}
						if(s[i+1][j+1] == 1&&turned == 0)
						{
							tmp[i+1][j+1] = 2;
							turned = 1;
						}
					}
					break;
				}
			case 3:
				{
					turned = 0;
					if(i == 0)
					{
						if(j == 0)
						{
							if(s[i][j+1] != 2&&s[i+1][j] != 2&&s[i+1][j+1] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
						else if(j == row-1)
						{
							if(s[i][j-1] != 2&&s[i+1][j-1] != 2&&s[i+1][j] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
						else
						{
							if(s[i][j-1] != 2&&s[i][j+1] != 2&&s[i+1][j-1] != 2&&s[i+1][j] != 2&&s[i+1][j+1] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
					}
					else if(i == col-1)
					{
						if(j == 0)
						{
							if(s[i-1][j] != 2&&s[i-1][j-1] != 2&&s[i][j+1] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
						else if(j == row-1)
						{
							if(s[i-1][j-1] != 2&&s[i-1][j] != 2&&s[i][j-1] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
						else
						{
							if(s[i-1][j-1] != 2&&s[i-1][j] != 2&&s[i-1][j+1] != 2&&s[i][j-1] != 2&&s[i][j+1] != 2&&turned == 0)
							{
								tmp[i][j] = 0;
								turned = 1;
							}
						}
					}
					else
					{
						if(s[i-1][j-1] != 2&&s[i-1][j] != 2&&s[i-1][j+1] != 2&&s[i][j-1] != 2&&s[i][j+1] != 2&&s[i+1][j-1] != 2&&s[i+1][j] != 2&&s[i+1][j+1] != 2)
						{
							tmp[i][j] = 0;
							turned = 1;
						}
					}

					if(s[i-1][j-1] == 2&&turned == 0)
					{
						tmp[i-1][j-1] = 3;
						turned = 1;
					}
					if(s[i-1][j] == 2&&turned == 0)
					{
						tmp[i-1][j] = 3;
						turned = 1;
					}
					if(s[i-1][j+1] == 2&&turned == 0)
					{
						tmp[i-1][j+1] = 3;
						turned = 1;
					}
					if(s[i][j-1] == 2&&turned == 0)
					{
						tmp[i][j-1] = 3;
						turned = 1;
					}
					if(s[i][j+1] == 2&&turned == 0)
					{
						tmp[i][j+1] = 3;
						turned = 1;
					}
					if(s[i+1][j-1] == 2&&turned == 0)
					{
						tmp[i+1][j-1] = 3;
						turned = 1;
					}
					if(s[i+1][j] == 2&&turned == 0)
					{
						tmp[i+1][j] = 3;
						turned = 1;
					}
					if(s[i+1][j+1] == 2&&turned == 0)
					{
						tmp[i+1][j+1] = 3;
						turned = 1;
					}
					break;
				}
			default:cout<<"error develop"<<endl;
		}
		for(i=0;i<col;i++)
			for(j=0;j<row;j++)
				if(tmp[i][j] != 5)
				{
					s[i][j] = tmp[i][j];
				}
        Traverse();
        Display();
}
int main()
{
	world me;
	me.Init();
	ShengTai your;
	your.Init();
	world *p = &me;
	p->Display();
	p = &your;
	p->Display();//通過基類指標呼叫派生類的虛擬函式

	ShengTai w;
	w.Init();
	int i;
	cout<<"input 1 to display,2 to develop,3 to exit."<<endl;
	cin>>i;
	while(i != 3)
	{
		if(i == 1)
			w.Display();
		if(i == 2)
			w.Develop();
		cout<<"input 1 to display,2 to develop,3 to exit."<<endl;
		cin>>i;
	}
	return 0;
}
 

執行結果：

1.     類的繼承與虛擬函式的實現

2.     生態系統中空地、青草、兔子、狐狸演化的過程

演化一次

演化二次