我正在尝试将3x3 tic-tac-toe算法转换为5x5 tic-tac-toe算法。我搜索了很多算法，但每次我修改代码时，计算机都从第一行开始，从该行的每一列继续。
这是我在网上找到的代码
公共课{
/* the board */
private int board[][];
/* empty */
public static final int EMPTY = 0;
/* player one */
public static final int ONE = 1;
/* player two */
public static final int TWO = 2;
public TicTacToeAI() {
board = new int[3][3];
}
/* get the board value for position (i,j) */
public int getBoardValue(int i,int j) {
if(i < 0 || i >= 3) return EMPTY;
if(j < 0 || j >= 3) return EMPTY;
return board[i][j];
}
/* set the board value for position (i,j) */
public void setBoardValue(int i,int j,int token) {
if(i < 0 || i >= 3) return;
if(j < 0 || j >= 3) return;
board[i][j] = token;
}
/* calculate the winning move for current token */
public int []nextWinningMove(int token) {
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
if(getBoardValue(i, j)==EMPTY) {
board[i][j] = token;
boolean win = isWin(token);
board[i][j] = EMPTY;
if(win) return new int[]{i,j};
}
return null;
}
public int inverse(int token) {
return token==ONE ? TWO : ONE;
}
/* calculate the best move for current token */
public int []nextMove(int token) {
/* lucky position in the center of board*/
if(getBoardValue(1, 1)==EMPTY) return new int[]{1,1};
/* if we can move on the next turn */
int winMove[] = nextWinningMove(token);
if(winMove!=null) return winMove;
/* choose the move that prevent enemy to win */
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
if(getBoardValue(i, j)==EMPTY)
{
board[i][j] = token;
boolean ok = nextWinningMove(inverse(token)) == null;
board[i][j] = EMPTY;
if(ok) return new int[]{i,j};
}
/* choose available move */
for(int i=0;i<3;i++)
for(int j=0;j<3;j++)
if(getBoardValue(i, j)==EMPTY)
return new int[]{i,j};
/* no move is available */
return null;
}
/* determine if current token is win or not win */
public boolean isWin(int token) {
final int DI[]={-1,0,1,1};
final int DJ[]={1,1,1,0};
for(int i=0;i<3;i++)
for(int j=0;j<3;j++) {
/* we skip if the token in position(i,j) not equal current token */
if(getBoardValue(i, j)!=token) continue;
for(int k=0;k<4;k++) {
int ctr = 0;
while(getBoardValue(i+DI[k]*ctr, j+DJ[k]*ctr)==token) ctr++;
if(ctr==3) return true;
}
}
return false;
}
}
这是我修改过的代码：
/* the board */
private int board[][];
/* empty */
public static final int EMPTY = 0;
/* player one */
public static final int ONE = 1;
/* player two */
public static final int TWO = 2;
public TicTacToeAI() {
board = new int[5][5];
}
/* get the board value for position (i,j) */
public int getBoardValue(int i,int j) {
if(i < 0 || i >= 5) return EMPTY;
if(j < 0 || j >= 5) return EMPTY;
return board[i][j];
}
/* set the board value for position (i,j) */
public void setBoardValue(int i,int j,int token) {
if(i < 0 || i >= 5) return;
if(j < 0 || j >= 5) return;
board[i][j] = token;
}
/* calculate the winning move for current token */
public int []nextWinningMove(int token) {
for(int i=0;i<5;i++)
for(int j=0;j<5;j++)
if(getBoardValue(i, j)==EMPTY) {
board[i][j] = token;
boolean win = isWin(token);
board[i][j] = EMPTY;
if(win) return new int[]{j,i};
}
return null;
}
public int inverse(int token) {
return token==ONE ? TWO : ONE;
}
/* calculate the best move for current token */
public int []nextMove(int token) {
/* lucky position in the center of board*/
if(getBoardValue(2, 2)==EMPTY) return new int[]{2,2};
/* if we can move on the next turn */
int winMove[] = nextWinningMove(token);
if(winMove!=null) return winMove;
/* choose the move that prevent enemy to win */
for(int i=0;i<5;i++)
for(int j=0;j<5;j++)
if(getBoardValue(i, j)==EMPTY)
{
board[i][j] = token;
boolean ok = nextWinningMove(inverse(token)) == null;
board[i][j] = EMPTY;
if(ok) return new int[]{i,j};
}
for(int i=1;i<4;i++)
for(int j=1;j<4;j++)
if(getBoardValue(i, j)==EMPTY)
return new int[]{i,j};
/* choose available move */
else{
for(i=0;i<5;i++)
for(j=0;j<5;j++)
if(getBoardValue(i, j)==EMPTY)
return new int[]{i,j};
}
/* no move is available */
return null;
}
/* determine if current token is win or not win */
public boolean isWin(int token) {
final int DI[]={-1,0,1,1,1,0};
final int DJ[]={1,1,1,0,-1,0};
for(int i=0;i<5;i++)
for(int j=0;j<5;j++) {
// we skip if the token in position(i,j) not equal current token
if(getBoardValue(i, j)!=token) continue;
for(int k=0;k<5;k++) {
int ctr = 0;
while(getBoardValue(i+DI[k]*ctr, j+DJ[k]*ctr)==token) ctr++;
if(ctr==4) return true;
}
}
return false;
}
}
谢谢你的帮助
最佳答案
原程序使用的策略只适用于3x3井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字井字（它可以完成一个链，如果它设法得到四个在一排，它会阻止一排的对手，但它不能做任何更复杂的事情）你需要设计一个新的战略，使这一工作。
关于java - 将3x3 tictactoe转换为5x5 tictactoe，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/16136065/