Unity3D实现简易五子棋源码

本文实例为大家分享了unity3d简易五子棋源码，供大家参考，具体内容如下

unity3d部分

对c#源码进行了改写简化：

using unityengine;
using system.collections;

public class chess : monobehaviour
{
  //四个锚点位置，用于计算棋子落点
  public gameobject lefttop;
  public gameobject righttop;
  public gameobject leftbottom;
  public gameobject rightbottom;
  //主摄像机
  public camera cam;
  //锚点在屏幕上的映射位置
  vector3 ltpos;
  vector3 rtpos;
  vector3 lbpos;
  vector3 rbpos;

  vector3 pointpos;//当前点选的位置
  float gridwidth = 1; //棋盘网格宽度
  float gridheight = 1; //棋盘网格高度
  float mingriddis; //网格宽和高中较小的一个
  vector2[,] chesspos; //存储棋盘上所有可以落子的位置
  int[,] chessstate; //存储棋盘位置上的落子状态
  enum turn { black, white };
  turn chessturn; //落子顺序
  public texture2d white; //白棋子
  public texture2d black; //黑棋子
  public texture2d blackwin; //白子获胜提示图
  public texture2d whitewin; //黑子获胜提示图
  int winner = 0; //获胜方，1为黑子，-1为白子
  bool isplaying = true; //是否处于对弈状态

  void start()
  {
    chesspos = new vector2[15, 15];
    chessstate = new int[17, 16];/*原来定义是new int[15, 15]，这里将原来数组chessstate上、下和右边各加一排数据，
    也就相当于在棋盘的上、下和右边各填加一排隐形的棋道。原因后面解释*/
    chessturn = turn.black;

    //计算锚点位置
    ltpos = cam.worldtoscreenpoint(lefttop.transform.position);
    rtpos = cam.worldtoscreenpoint(righttop.transform.position);
    lbpos = cam.worldtoscreenpoint(leftbottom.transform.position);
    rbpos = cam.worldtoscreenpoint(rightbottom.transform.position);

    //计算网格宽度
    gridwidth = (rtpos.x - ltpos.x) / 14;
    gridheight = (ltpos.y - lbpos.y) / 14;
    mingriddis = gridwidth < gridheight ? gridwidth : gridheight;

    //计算落子点位置
    for (int i = 0; i < 15; i++)
    {
      for (int j = 0; j < 15; j++)
      {
        chesspos[i, j] = new vector2(lbpos.x + gridwidth * j, lbpos.y + gridheight * i);//这里和源程序定义稍有不同，这里i定位行，j为列
      }
    }
  }

  void update()
  {
    //检测鼠标输入并确定落子状态
    if (isplaying && input.getmousebuttondown(0))
    {
      pointpos = input.mouseposition;
      for (int i = 0; i < 15; i++)
      {
        for (int j = 0; j < 15; j++)
        {
          //找到最接近鼠标点击位置的落子点，如果空则落子
          if (dis(pointpos, chesspos[i, j]) < mingriddis / 2 && chessstate[i + 1, j] == 0)/*这里chessstate行要加1，
            因为上、下和右边各多加了一排，要空出来，chesspos的i行对应chessstate的i+1行*/
          {
            //根据下棋顺序确定落子颜色
            chessstate[i + 1, j] = chessturn == turn.black ? 1 : -1;//同理
            //落子成功，更换下棋顺序
            chessturn = chessturn == turn.black ? turn.white : turn.black;
          }
        }
      }
      //调用判断函数，确定是否有获胜方
      int re = result();
      if (re == 1)
      {
        debug.log("黑棋胜");
        winner = 1;
        isplaying = false;
      }
      else if (re == -1)
      {
        debug.log("白棋胜");
        winner = -1;
        isplaying = false;
      }
    }
    //按下空格重新开始游戏
    if (input.getkeydown(keycode.space))
    {
      for (int i = 0; i < 15; i++)
      {
        for (int j = 0; j < 15; j++)
        {
          chessstate[i + 1, j] = 0;//同理
        }
      }
      isplaying = true;
      chessturn = turn.black;
      winner = 0;
    }
  }
  //计算平面距离函数
  float dis(vector3 mpos, vector2 gridpos)
  {
    return mathf.sqrt(mathf.pow(mpos.x - gridpos.x, 2) + mathf.pow(mpos.y - gridpos.y, 2));
  }

  void ongui()
  {
    //绘制棋子
    for (int i = 0; i < 15; i++)
    {
      for (int j = 0; j < 15; j++)
      {
        if (chessstate[i + 1, j] == 1)//同理
        {
          gui.drawtexture(new rect(chesspos[i, j].x - gridwidth / 2, screen.height - chesspos[i, j].y - gridheight / 2, gridwidth, gridheight), black);
        }
        if (chessstate[i + 1, j] == -1)//同理
        {
          gui.drawtexture(new rect(chesspos[i, j].x - gridwidth / 2, screen.height - chesspos[i, j].y - gridheight / 2, gridwidth, gridheight), white);
        }
      }
    }
    //根据获胜状态，弹出相应的胜利图片
    if (winner == 1)
    {
      gui.drawtexture(new rect(screen.width * 0.25f, screen.height * 0.25f, screen.width * 0.5f, screen.height * 0.25f), blackwin);
    }
    if (winner == -1)
      gui.drawtexture(new rect(screen.width * 0.25f, screen.height * 0.25f, screen.width * 0.5f, screen.height * 0.25f), whitewin);
  }
 //改写result函数
 /*解释：c语言中，这样的表达式：chessstate[i]&&chessstate[i+1]&&chessstate[i+2]&&chessstate[i+3]&&chessstate[i+4]，如果
   * chessstate[i]为false,则不管b是真是假或者是异常都不会运行，利用这一点，在chessstate的右边、上边和下边各加一行为0的数据，
   * 这样在判断连续五个棋子的状态时，就不用担心chessstate数组的索引值超出范围。例如：chessstate[i+4]的索引值i+4刚好超出范围，
   * 通过在原来数组chessstate的上、下和右边个添加一排为0的数，这样chessstate[i+3]==0，于是就可以避免引起异常，从而简化代码*/
  int result()
  {
    int flag = 0;
    if (chessturn == turn.white)
    {
      for (int i = 1; i <= 15; i++)//这里的i从1开始
      {
        for (int j = 0; j <= 14; j++)//j不用变
        {
          if ((chessstate[i, j] == 1 && chessstate[i, j + 1] == 1 && chessstate[i, j + 2] == 1 && chessstate[i, j + 3] == 1 && chessstate[i, j + 4] == 1)//向右横向
            || (chessstate[i, j] == 1 && chessstate[i + 1, j] == 1 && chessstate[i + 2, j] == 1 && chessstate[i + 3, j] == 1 && chessstate[i + 4, j] == 1)//向上横向
            || (chessstate[i, j] == 1 && chessstate[i + 1, j + 1] == 1 && chessstate[i + 2, j + 2] == 1 && chessstate[i + 3, j + 3] == 1 && chessstate[i + 4, j + 4] == 1)//向右上斜向
            || (chessstate[i, j] == 1 && chessstate[i - 1, j + 1] == 1 && chessstate[i - 2, j + 2] == 1 && chessstate[i - 3, j + 3] == 1 && chessstate[i - 4, j + 4] == 1))//向右下斜向
          {
            flag = 1;
          }
        }
      }
    }
    else if (chessturn == turn.black)
    {
      for (int i = 1; i <= 15; i++)//这里的i从1开始
      {
        for (int j = 0; j <= 14; j++)
        {

          if ((chessstate[i, j] == -1 && chessstate[i, j + 1] == -1 && chessstate[i, j + 2] == -1 && chessstate[i, j + 3] == -1 && chessstate[i, j + 4] == -1)
            || (chessstate[i, j] == -1 && chessstate[i + 1, j] == -1 && chessstate[i + 2, j] == -1 && chessstate[i + 3, j] == -1 && chessstate[i + 4, j] == -1)
            || (chessstate[i, j] == -1 && chessstate[i + 1, j + 1] == -1 && chessstate[i + 2, j + 2] == -1 && chessstate[i + 3, j + 3] == -1 && chessstate[i + 4, j + 4] == -1)
            || (chessstate[i, j] == -1 && chessstate[i - 1, j + 1] == -1 && chessstate[i - 2, j + 2] == -1 && chessstate[i - 3, j + 3] == -1 && chessstate[i - 4, j + 4] == -1))
          {
            flag = -1;
          }
        }
      }
    }
    return flag;
  }
}

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。