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這篇文章主要介紹C++怎么實現趣味掃雷游戲,文中介紹的非常詳細,具有一定的參考價值,感興趣的小伙伴們一定要看完!
1.初始化陣列。
2.輸入坐標點。
3.選擇:挖掘,標記,取消標記,重啟,退出游戲。
如果選了挖掘,判斷坐標點是地雷則游戲結束,是數字則顯示數字并回到2,是空格則顯示周圍8個元素值并直到連帶的空格顯示完了回到2;
如果選了標記,將該點的元素值設為-2并回到2;
如果選了取消標記,初始化該點,回到2;
如果選了重啟,則初始化陣列,回到2;
如果選了退出游戲,則exit。
4.挖掘完所有非地雷點后,游戲勝利,選擇是否再來一局,是則回到1,否則exit
創建一個bombsweep類,存儲幾個方法:
calculate:統計以(x,y)為中心周圍8個點的地雷數目。
game:模擬游戲過程。
print:打印陣列。
check:檢查是否滿足勝利條件。
在main函數中,在需要的時候根據bombsweep類創建bs對象,調用bs里面的相關方法。
程序代碼
#include <ctime> #include <cstdlib> #include <iostream> #include <cstring> using namespace std; int map[12][12]; // ??????????,????????????1 int derection[3] = { 0, 1, -1 }; //????????8????? int type; class bombsweep { public: int calculate ( int x, int y ) { int counter = 0; for ( int i = 0; i < 3; i++ ) for ( int j = 0; j < 3; j++ ) if ( map[ x+derection[i]][ y+derection[j] ] == 9 ) counter++; // ???(x,y)?????8??????? return counter; } void game ( int x, int y ) { if ( calculate ( x, y ) == 0 ) { map[x][y] = 0; for ( int i = 0; i < 3; i++ ) { // ???????,????????? for ( int j = 0; j < 3; j++ ) if ( x+derection[i] <= 9 && y+derection[j] <= 9 && x+derection[i] >= 1 && y+derection[j] >= 1 && !( derection[i] == 0 && derection[j] == 0 ) && map[x+derection[i]][y+derection[j]] == -1 ) game( x+derection[i], y+derection[j] ); // ???????????????0,????????! } //????????.??????????? } else map[x][y] = calculate(x,y); } void print (int x,int y) { cout << " |"; for (int i=1; i<10; i++) cout << " " << i; cout << endl; cout << "__|__________________Y" ; cout << endl; for ( int i = 1; i < 10; i++ ) { cout << i << " |"; for ( int j = 1; j < 10; j++ ) { if(map[i][j]==-2) cout <<" B"; else if ( map[i][j] == -1 || map[i][j] == 9 ) cout << " #"; else cout << " "<< map[i][j]; } cout << "\n"; } cout << " X\n"; } bool check () { int counter = 0; for ( int i = 1; i < 10; i++ ) for ( int j = 1; j < 10; j++ ) if ( map[i][j] != -1 ) counter++; if ( counter == 10 ) return true; else return false; } }; int main () { int i, j, x, y; char ch; srand ( time ( 0 ) ); do { //????? memset ( map, -1, sizeof(map) ); for ( i = 0; i < 10; ) { x = rand()%9 + 1; y = rand()%9 + 1; if ( map[x][y] != 9 ) { map[x][y] = 9; i++; } } cout << " |"; for (i=1; i<10; i++) cout << " " << i; cout << endl; cout << "__|__________________Y" ; cout << endl; for ( i = 1; i < 10; i++ ) { cout << i << " |"; for ( j = 1; j < 10; j++ ) cout << " "<< "#"; cout << "\n"; } cout << " X\n"; cout << "Please input location x,press enter then input location y: \n"; while ( cin >> x >> y ) { cout << "Please select:1.dig, 2.sign, 3.cancel sign, 4.restart, 5.exit: \n"; cin >>type; switch(type) { case 1: { if ( map[x][y] == 9 || map[x][y]==-2) { cout << "YOU LOSE!" << endl; cout << " |"; for (i=1; i<10; i++) cout << " " << i; cout << endl; cout << "__|__________________Y"<<endl ; for ( i = 1; i < 10; i++ ) { cout << i << " |"; for ( j = 1; j < 10; j++ ) { if ( map[i][j] == 9 || map[i][j]==-2) cout << " @"; else cout << " #"; } cout << "\n"; } cout << " X\n"; exit(0); } bombsweep bs; bs.game(x,y); bs.print(x,y); cout << "Please input location x,press enter then input location y: \n"; if ( bs.check()) { cout << "YOU WIN" << endl; break; } continue; } case 2: { bombsweep bs; map[x][y]=-2; bs.print(x,y); cout << "Please input location x,press enter then input location y: \n"; continue; } case 3: { bombsweep bs; map[x][y]=-1; bs.print(x,y); cout << "Please input location x,press enter then input location y: \n"; continue; } case 4: { memset ( map, -1, sizeof(map) ); for ( i = 0; i < 10; ) { x = rand()%9 + 1; y = rand()%9 + 1; if ( map[x][y] != 9 ) { map[x][y] = 9; i++; } } cout << " |"; for (i=1; i<10; i++) cout << " " << i; cout << endl; cout << "__|__________________Y" ; cout << endl; for ( i = 1; i < 10; i++ ) { cout << i << " |"; for ( j = 1; j < 10; j++ ) cout << " "<< "#"; cout << "\n"; } cout << " X\n"; cout << "Please input location x,press enter then input location y: \n"; continue; } case 5: cout << "Game Ended\n"; exit(0); break; default: cout<< "Invalid input, try again: \n"; continue; }//end switch }//end while(cin >> x >>y) cout << "Do you want to play again?(y/n):" << endl; cin >> ch; }//end do while ( ch == 'y' ); return 0; }//end main()
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