20141026--娱乐-箱子

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace XTU0
{
    class Program
    {
        #region===将数值型的二维数组的值，赋值给字符串型的二维数组，并打印地图===
        /// <summary>
        /// 将数值型的二维数组的值，赋值给字符串型的二维数组tu，并打印地图
        /// </summary>
        /// <param name="tu"></param>
        public void ditus(int[,,] tu,int m)
        {
            string[,] ditu = new string[10, 11];
            for (int i = 0; i < 10; i++)
            {
                for (int j = 0; j < 10; j++)
                {
                    if (tu[m, i, j] == 0) { ditu[i, j] = "  "; }
                    if (tu[m, i, j] == 1) { ditu[i, j] = "■"; }
                    if (tu[m, i, j] == 2) { ditu[i, j] = "●"; }
                    if (tu[m, i, j] == 3) { ditu[i, j] = "◎"; }
                    if (tu[m, i, j] == 4) { ditu[i, j] = "人"; }
                    if (tu[m, i, j] == 5) { ditu[i, j] = "※"; }
                }
                ditu[i, 10] = "
";
            }
            foreach (string s in ditu)//打印出地图
            {
                Console.Write(s);
            }
        }
        #endregion

        static void Main(string[] args)
        {
            const int guan=3;//const  定义常量，作为有多少关
            Program pd = new Program();//初始化函数
            int[,,] tu = new int[guan,10, 10]
            {
                //0表示空地，1表示墙，2表示箱子xzx,xzy，3表示箱子空位kwx,kwy，4表示光标gbx,gby，5表示放好的箱子fhx,fhy
                {//7表示空位上的光标，8表示钥匙ysx,ysy，9表示带锁的出口ckx,cky
                {1,1,1,1,1,1,1,1,1,1},//j,y横坐标0-9
                {1,0,0,1,0,0,0,0,0,1},//纵坐标i，x
                {1,4,0,0,0,0,2,0,3,1},//0
                {1,0,1,1,0,1,0,0,0,1},// |
                {1,0,0,0,0,0,0,0,0,1},//9
                {1,3,1,1,1,1,1,2,1,1},
                {1,2,1,0,0,0,0,0,0,1},
                {1,0,0,0,0,1,1,0,0,1},
                {1,1,0,1,0,0,0,3,0,1},
                {1,1,1,1,1,1,1,1,1,1},
                },
                {//第二关
                {1,1,1,1,1,1,1,1,1,1},
                {1,1,1,1,4,0,1,0,3,1},
                {1,1,1,3,0,0,2,0,0,1},
                {1,1,0,0,2,0,1,0,0,1},
                {1,0,0,0,1,1,1,0,0,1},
                {1,0,1,1,0,0,0,0,0,1},
                {1,0,0,0,0,2,0,0,1,1},
                {1,0,3,0,0,0,1,1,1,1},
                {1,0,0,0,0,1,1,1,1,1},
                {1,1,1,1,1,1,1,1,1,1},
                },
                {//第3关
                {1,1,1,1,1,1,1,1,1,1},
                {1,1,0,0,4,0,0,3,1,1},
                {1,1,0,2,0,0,0,2,3,1},
                {1,1,0,0,2,1,1,0,0,1},
                {1,1,0,1,0,0,0,3,0,1},
                {1,0,3,0,0,0,1,0,1,1},
                {1,0,0,1,1,2,0,0,1,1},
                {1,3,2,0,0,0,2,0,1,1},
                {1,1,3,0,0,0,0,0,1,1},
                {1,1,1,1,1,1,1,1,1,1},
                },
            };
            for (int m = 0; m < guan; )
            {
            #region===从地图上获得互动元素的坐标，并打印地图===
            int x = 0, y = 0;/*xy表示光标坐标  4*/
            //int ysx = 0, ysy = 0;/*表示钥匙的坐标  8*/
            //int ckx = 0, cky = 0;/*表示出口坐标   9*/
            int xzx = 0, xzy = 0;/*表示箱子坐标  2*/
            int kwx = 0, kwy = 0;/*表示箱子空 3*/
            int fhx = 0, fhy = 0;/*表示放好的箱子 5*/
            int yao = 0, shi = 0;
                for (int i = 0; i < 10; i++)//从地图上获得互动元素的坐标
                {
                    for (int j = 0; j < 10; j++)
                    {
                        if (tu[m, i, j] == 2) { xzx = i; xzy = j; }
                        if (tu[m, i, j] == 3) { kwx = i; kwy = j; shi++;}
                        if (tu[m, i, j] == 4) { x = i; y = j; }
                        if (tu[m, i, j] == 5) { fhx = i; fhy = j; }
                        //if (tu[m, i, j] == 8) { ysx = i; ysy = j; }
                        //if (tu[m, i, j] == 9) { ckx = i; cky = j; }
                    }
                }
                pd.ditus(tu, m);
                
            #endregion
                string jilu="  ";
                int tui = 0;
                for (; ; )
                {
                    if (yao == shi)
                    {
                        m = m + 1;
                        Console.WriteLine("过关！！
按回车进入下一关。");
                        Console.ReadLine();
                        Console.Clear();
                        break;
                    }
                    Console.WriteLine("按w,a,s,d移动
推错时按backspace后退一步");
                    ConsoleKeyInfo fangxiang = Console.ReadKey();
                    string shuru = fangxiang.Key.ToString().ToLower();
                    if(shuru=="enter")
                    {
                        Console.WriteLine("输入next进入下一关");
                        string ne= Console.ReadLine();
                        if (ne == "next") { m++; Console.Clear(); break; }
                    }
                    if (shuru == "backspace"&&tui==1)
                    {
                        if (tu[m, x, y - 1] == 0)
                        {
                            if (jilu == "d0") { tu[m, x, y] = 2; tu[m, x, y + 1] = 0; tu[m, x, y - 1] = 4; y = y - 1; tui = 0; }
                            if (jilu == "d1") { tu[m, x, y] = 2; tu[m, x, y + 1] = 3; tu[m, x, y - 1] = 4; y = y - 1; yao--; tui = 0; }
                        }
                        if (tu[m, x, y + 1] == 0)
                        {
                            if (jilu == "a0") { tu[m, x, y] = 2; tu[m, x, y - 1] = 0; tu[m, x, y + 1] = 4; y = y + 1; tui = 0; }
                            if (jilu == "a1") { tu[m, x, y] = 2; tu[m, x, y - 1] = 3; tu[m, x, y + 1] = 4; y = y + 1; yao--; tui = 0; }
                        }
                        if (tu[m, x + 1, y] == 0)
                        {
                            if (jilu == "w0") { tu[m, x, y] = 2; tu[m, x - 1, y] = 0; tu[m, x + 1, y] = 4; x = x + 1; tui = 0; }
                            if (jilu == "w1") { tu[m, x, y] = 2; tu[m, x - 1, y] = 3; tu[m, x + 1, y] = 4; x = x + 1; yao--; tui = 0; }
                        }
                        if (tu[m, x - 1, y] == 0)
                        {
                            if (jilu == "s0") { tu[m, x, y] = 2; tu[m, x + 1, y] = 0; tu[m, x - 1, y] = 4; x = x - 1; tui = 0; }
                            if (jilu == "s1") { tu[m, x, y] = 2; tu[m, x + 1, y] = 3; tu[m, x - 1, y] = 4; x = x - 1; yao--; tui = 0; }
                        }
                        Console.Clear();
                        pd.ditus(tu, m);
                        Console.WriteLine("每次只能退一步");
                    }
                    #region===d 向右===
                    if (shuru == "d" && tu[m, x, y + 1] != 1 && tu[m, x, y + 1] != 3 && tu[m, x, y + 1] != 5)//如果光标下一步不是墙，进入if判断下一步是什么
                    {
                        //if (tu[x, y] == 7) { }
                        if (tu[m, x, y + 1] == 2 && tu[m, x, y + 2] != 1 && tu[m, x, y + 2] != 2 && tu[m, x, y + 2] != 5)//光标的下一步是箱子，并且箱子的下一步不是墙，不是箱子，不是放好的箱子
                        {
                            if (tu[m, x, y + 2] == 0)//箱子的下一步是空地
                            {
                                tu[m, x, y + 2] = 2;//箱子的下一步变成箱子
                                jilu = "d0"; tui = 1;
                            }
                            if (tu[m, x, y + 2] == 3)//如果箱子下一步是空位
                            {
                                tu[m, x, y + 2] = 5;//箱子下一步变成放好的箱子
                                yao++;
                                jilu = "d1"; tui = 1;
                            } 
                            tu[m, x, y + 1] = 4;
                            tu[m, x, y] = 0;
                            y = y + 1;
                        }
                        if (tu[m, x, y + 1] == 0)
                        {
                            tu[m, x, y + 1] = 4;
                            tu[m, x, y] = 0;
                            y = y + 1;
                            jilu = "  ";
                        }
                            Console.Clear();
                            pd.ditus(tu, m);
                    }
                    #endregion
                    #region===a 向左===
                    if (shuru == "a" && tu[m, x, y - 1] != 1 && tu[m, x, y - 1] != 3 && tu[m, x, y - 1] != 5)//如果光标下一步不是墙，进入if判断下一步是什么
                    {
                        if (tu[m, x, y - 1] == 2 && tu[m, x, y - 2] != 1 && tu[m, x, y - 2] != 2 && tu[m, x, y - 2] != 5)//光标的下一步是箱子，并且箱子的下一步不是墙，不是箱子，不是放好的箱子
                        {
                            if (tu[m, x, y - 2] == 0)//箱子的下一步是空地
                            {
                                tu[m, x, y - 2] = 2;//箱子的下一步变成箱子
                                jilu = "a0"; tui = 1;
                            }
                            if (tu[m, x, y - 2] == 3)//如果箱子下一步是空位
                            {
                                tu[m, x, y - 2] = 5;//箱子下一步变成放好的箱子
                                yao++;
                                jilu = "a1"; tui = 1;
                            }
                            tu[m, x, y] = 0;//光标变成空地
                            tu[m, x, y - 1] = 4;//光标的下一步变成光标
                            y = y - 1;//光标坐标改变
                        }
                        if (tu[m, x, y - 1] == 0)
                        {
                            tu[m, x, y] = 0;//光标变成空地
                            tu[m, x, y - 1] = 4;//光标的下一步变成光标
                            y = y - 1;//光标坐标改变
                            jilu = "  ";
                        }
                            Console.Clear();
                            pd.ditus(tu, m);
                    }
                    #endregion
                    #region===w 向上===
                    if (shuru == "w" && tu[m, x - 1, y] != 1 && tu[m, x - 1, y] != 3 && tu[m, x - 1, y] != 5)//如果光标下一步不是墙，进入if判断下一步是什么
                    {
                        if (tu[m, x - 1, y] == 2 && tu[m, x - 2, y] != 1 && tu[m, x - 2, y] != 2 && tu[m, x - 2, y] != 5)//光标的下一步是箱子，并且箱子的下一步不是墙，不是箱子，不是放好的箱子
                        {
                            if (tu[m, x - 2, y] == 0)//箱子的下一步是空地
                            {
                                tu[m, x - 2, y] = 2;//箱子的下一步变成箱子
                                jilu = "w0"; tui = 1;
                            }
                            if (tu[m, x - 2, y] == 3)//如果箱子下一步是空位
                            {
                                tu[m, x - 2, y] = 5;//箱子下一步变成放好的箱子
                                yao++;
                                jilu = "w1"; tui = 1;
                            }
                            tu[m, x, y] = 0;
                            tu[m, x - 1, y] = 4;
                            x = x - 1;
                        }
                        if (tu[m, x - 1, y] == 0)
                        {
                            tu[m, x, y] = 0;
                            tu[m, x - 1, y] = 4;
                            x = x - 1;
                            jilu = "  ";
                        }
                            Console.Clear();
                            pd.ditus(tu, m);
                    }
                    #endregion
                    #region===s 向下===
                    if (shuru == "s" && tu[m, x + 1, y] != 1 && tu[m, x + 1, y] != 3&& tu[m, x + 1, y] != 5)//如果光标下一步不是墙，进入if判断下一步是什么
                    {
                        if (tu[m, x + 1, y] == 2 && tu[m, x + 2, y] != 1 && tu[m, x + 2, y] != 2 && tu[m, x + 2, y] != 5)//光标的下一步是箱子，并且箱子的下一步不是墙，不是箱子，不是放好的箱子
                        {
                            if (tu[m, x + 2, y] == 0)//箱子的下一步是空地
                            {
                                tu[m, x + 2, y] = 2;//箱子的下一步变成箱子
                                jilu = "s0"; tui = 1;
                            }
                            if (tu[m, x + 2, y] == 3)//如果箱子下一步是空位
                            {
                                tu[m, x + 2, y] = 5;//箱子下一步变成放好的箱子
                                yao++;
                                jilu = "s1"; tui = 1;
                            }
                            tu[m, x, y] = 0;//光标变成空地
                            tu[m, x + 1, y] = 4;//光标的下一步变成光标
                            x = x + 1;//光标坐标改变
                        }
                        if (tu[m, x + 1, y] == 0)
                        {
                            tu[m, x, y] = 0;//光标变成空地
                            tu[m, x + 1, y] = 4;//光标的下一步变成光标
                            x = x + 1;//光标坐标改变
                            jilu = "  ";
                        }
                        Console.Clear();
                        pd.ditus(tu, m);
                    }
                    #endregion
                }
            }
            Console.ReadLine();
        }
    }
}