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;
  }
}

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