教你怎麼用c++基本語法實現一個簡單的五子棋小遊戲
2020-09-28 09:01:47
這個小遊戲是在2020年5月份無聊寫的，程式碼量不大，權當娛樂哈
基本思路：
1.建立一個15*15棋盤類，並設計相關函數（輸出棋盤，下黑棋，下白棋等）
2.編寫judge()函數，判斷勝負條件
3.主函數思路：黑方先下，每次不斷更新棋盤的元素：如果該處已經有棋子了，則重新下一次，以此不斷迴圈直到遊戲結束。
程式碼如下：
1.標頭檔案
//wuziqi.h
#ifndef _WUZIQI_H_
#define _WUZIQI_H_
#include <iostream>
using namespace std;
class chessboard;
class chessboard
{
public:
int judge();
void build();//建立棋盤
void changeblack(int goalx, int goaly, int& goal);//改變棋盤陣列元素即更新棋子
void changewhite(int goalx, int goaly, int& goal);
void out();//輸出棋盤
private:
string arr[15][15];
};
#endif // !_WUZIQI_H_
2.cpp檔案
//wuziqi.cpp
#include <iostream>
#include "wuziqi.h"
using namespace std;
int chessboard::judge()
{
for (int i = 0; i < 11; i++)
{
for (int j = 0; j < 11; j++)
{
if (//橫一樣
arr[i][j] == "●" &&
arr[i][j + 1] == "●" &&
arr[i][j + 2] == "●" &&
arr[i][j + 3] == "●" &&
arr[i][j + 4] == "●"
) return 0;//不能使用連等
}
for (int j = 0; j < 11; j++)
{
if (//豎一樣
arr[j][i] == "●" &&
arr[j + 1][i] == "●" &&
arr[j + 2][i] == "●" &&
arr[j + 3][i] == "●" &&
arr[j + 4][i] == "●"
) return 0;
}
for (int j = 0; j < 11; j++)
{
if (//順斜一樣
arr[i][j] == "●" &&
arr[i + 1][j + 1] == "●" &&
arr[i + 2][j + 2] == "●" &&
arr[i + 3][j + 3] == "●" &&
arr[i + 4][j + 4] == "●"
) return 0;
}
for (int j = 14; j > 3; j--)
{
if (//逆斜一樣
arr[i][j] == "●" &&
arr[i + 1][j - 1] == "●" &&
arr[i + 2][j - 2] == "●" &&
arr[i + 3][j - 3] == "●" &&
arr[i + 4][j - 4] == "●"
) return 0;
}
for (int j = 0; j < 11; j++)
{
if (//橫一樣
arr[i][j] == "○" &&
arr[i][j + 1] == "○" &&
arr[i][j + 2] == "○" &&
arr[i][j + 3] == "○" &&
arr[i][j + 4] == "○"
) return 0;//不能使用連等
}
for (int j = 0; j < 11; j++)
{
if (//豎一樣
arr[j][i] == "○" &&
arr[j + 1][i] == "○" &&
arr[j + 2][i] == "○" &&
arr[j + 3][i] == "○" &&
arr[j + 4][i] == "○"
) return 0;
}
for (int j = 0; j < 11; j++)
{
if (//順斜一樣//列從0開始
arr[i][j] == "○" &&
arr[i + 1][j + 1] == "○" &&
arr[i + 2][j + 2] == "○" &&
arr[i + 3][j + 3] == "○" &&
arr[i + 4][j + 4] == "○"
) return 0;
}
for (int j = 14; j > 3; j--)
{
if (//逆斜一樣//列從14開始
arr[i][j] == "○" &&
arr[i + 1][j - 1] == "○" &&
arr[i + 2][j - 2] == "○" &&
arr[i + 3][j - 3] == "○" &&
arr[i + 4][j - 4] == "○"
) return 0;
}
}
return 1;
}
//棋盤--------------------------------------------------------------------------------------
void chessboard::build()
{
for (int i = 0; i < 15; i++)
{
for (int j = 0; j < 15; j++)
{
arr[i][j] = " ";
}
}//初始化棋盤
}
void chessboard::changeblack(int goalx, int goaly, int& goal)//更新點位
{
if (arr[goalx][goaly] != " ")
{
cout << "這一點已經有子了,請重新下一次黑棋" << endl;
goal--;
}
else
arr[goalx][goaly] = "○";
}
void chessboard::changewhite(int goalx, int goaly, int& goal)//更新點位
{
if (arr[goalx][goaly] != " ")
{
cout << "這一點已經有子了,請重新下一次白旗" << endl;
goal--;
}
else
arr[goalx][goaly] = "●";
}
void chessboard::out()
{
cout << " 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 " << endl;
for (int i = 0; i < 15; i++)
{
cout << "---------------------------------------------------------------" << endl;
if (i < 10)
cout << i << " | ";
else
cout << i << "| ";
for (int j = 0; j < 15; j++)
{
cout << arr[i][j] << "| ";
}
cout << endl;
}//輸出棋盤，每次輸入一個點就輸出一次棋盤
cout << "---------------------------------------------------------------" << endl;
}
//main.cpp
#include <iostream>
#include "wuziqi.h"
using namespace std;
int main(void)
{
chessboard goalboard;//目標棋盤
goalboard.build();
goalboard.out();
for (int i = 1; i < 10000; i++)
{
if (i % 2 != 0)//奇數黑方走，偶數白方走
{
int x = 0, y = 0;
cout << "請輸入黑棋x座標和y座標" << endl;
cin >> y >> x;
goalboard.changeblack(x, y, i);//持續更新黑棋盤
goalboard.out();
if (goalboard.judge() == 0)
{
cout << "遊戲結束 黑棋勝利！" << endl;
cout << "輸入任意座標重新開始" << endl;
goalboard.build();
}
}
else
{
int x = 0, y = 0;
cout << "請輸入白棋x座標和y座標" << endl;
cin >> y >> x;
goalboard.changewhite(x, y, i);//持續更新白棋盤
goalboard.out();
if (goalboard.judge() == 0)
{
cout << "遊戲結束 白旗勝利！" << endl;
cout << "輸入任意座標重新開始" << endl;
goalboard.build();
}
}
}
return 0;
}
執行效果圖：
PS：
這個小遊戲還有不少不足之處，如輸入棋子座標超過棋盤大小怎麼辦？勝負的判定是否還有更簡單的方法？都值得再去思考完善。
菜雞一枚難免有很多不足之處，歡迎指教！感激不盡！