我已经使用C ++进行了很长时间的尝试,以尝试构建一个稍微简单的井字游戏。玩家轮流提供所需元素,将1和2加到动态尺寸的游戏板上。
用来确定玩家的回合和游戏板更新的功能正常,但是下面的代码检查0所代表的板的获胜条件,例如:
但是不起作用。经过测试后,该板仅需每轮更新一次。我怀疑的问题是数组算法。如何执行该算法以成功检测条件?提前致谢。
bool chkCondition(int rows, int columns, int**board){
int counter=0;
for(int i=0; i<rows; i++){
for(int j=0; j<columns; j++){
if(board[i][j]==1 && board[i][j]+1==1 && board[i][j]+2==1){//checks horizontally for 1
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==1 && board[i][j]+columns==1 && board[i][j]+2*columns==1){//checks vertically for 1
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==1 && board[i][j]+columns+1==1 && board[i][j]+2*columns+2==1){//checks diagonally for 1
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j]+1==2 && board[i][j]+2==2){//checks horizontally for 2
cout << "Player 2 wins!";
return true;
}else if(board[i][j]==2 && board[i][j]+columns==2 && board[i][j]+2*columns==2){//checks vertically for 2
cout << "Player 2 wins!";
return true;
}else if(board[i][j]==2 && board[i][j]+columns+1==2 && board[i][j]+2*columns+2==2){//checks diagonally for 2
cout << "Player 2 wins!";
return true;
}else if(board[i][j]==0){//attempts to detect if the game is a draw
if(counter==(rows+columns)-3){
cout << "It's a draw!";
return true;
}
}
}
}
return false;
counter++;
}
编辑将算术更改为正确的格式,以查找元素而不是添加其值。新代码在这里:
bool chkCondition(int rows, int columns, int**board){
int counter=0;
for(int i=0; i<rows; i++){
counter++;
for(int j=0; j<columns; j++){
if(board[i][j]==1 && board[i][j+1]==1 && board[i][j+2]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j+1]==2 && board[i][j+2]==2){
cout << "Player 2 wins!";
return true;
}
if(counter==rows | counter>rows-1){
//cout << "L--------\n";
//cout << "checking vertical\n";
if(board[i][j]==1 && board[i][j-columns]==1 && board[i][j-columns-columns]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j-columns]==2 && board[i][j-columns-columns]==2){
cout << "Player 2 wins!";
return true;
//cout << "checking diagonal\n";
}else if(board[i][j]==1 && board[i][j-columns-1]==1 && board[i][j-columns-columns-2]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j-columns-1]==2 && board[i][j-columns-columns-2]==2){
cout << "Player 2 wins!";
return true;
}
}
if(counter==1 && counter<rows-1){
//cout << "H--------\n";
//cout << "checking vertical\n";
if(board[i][j]==1 && board[i][j+columns]==1 && board[i][j+columns+columns]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j+columns]==2 && board[i][j+columns+columns]==2){
cout << "Player 2 wins!";
return true;
//cout << "checking diagonal\n";
}else if(board[i][j]==1 && board[i][j+columns+1]==1 && board[i][j+columns+columns+2]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j+columns+1]==2 && board[i][j+columns+columns+2]==2){
cout << "Player 2 wins!";
return true;
}
}
if(counter>2 && counter<rows-1){
//cout << "M--------\n";
//cout << "checking vertical\n";
if(board[i][j]==1 && board[i][j+columns]==1 && board[i][j-columns]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j+columns]==2 && board[i][j-columns]==2){
cout << "Player 2 wins!";
return true;
//cout << "checking diagonal\n";
}else if(board[i][j]==1 && board[i][j+columns+1]==1 && board[i][j-columns-1]==1){
cout << "Player 1 wins!";
return true;
}else if(board[i][j]==2 && board[i][j+columns+1]==2 && board[i][j-columns-1]==2){
cout << "Player 2 wins!";
return true;
}
}
}
}
return false;
}
外部if语句用于确定在迭代中对当前行进行计数的计数器是在表的开头,中间还是结尾,以避免分段错误。这可以正常工作,并且程序不会崩溃,但是,尽管算术是逻辑上的,但所有垂直和对角条件语句根本不起作用。谁能告诉我发生了什么事?
正如评论所说,问题在于您如何访问表。另一个问题是如何使用循环。如果您首先进行水平检查,则循环遍历每一行,但不应遍历每一列,否则您会遇到很好的分段错误。例如第一部分(假设您实际上将+ 1 / + 2放在了合适的位置
for(int i=0; i<rows; i++){
for(int j=0; j<columns; j++){
if(board[i][j]==1 && board[i][j+1]==1 && board[i][j+2]==1){//checks horizontally for 1
cout << "Player 1 wins!";
return true;
当j为column-2或column时,它将给您带来分段错误。为了解决这个问题,您需要先停下来。对于水平检查,线条相同。只执行这样的循环会更复杂,否则您需要添加更多ifs来避免此错误。
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句