Realization of simple maze game _c language based on C language

This article describes the C language to achieve a simple maze game method, code integrity, easy to understand the reader.

Learn the data structure of the "stack" written by a maze of procedures, actually used to two-way queues, convenient after the operation of the output after the point.

#include <cstdio> #include <deque> #include <windows.h> using namespace std;
  Class Node {public:int x,y;
int lastopt;
};
Deque<node> STA;
int x,y;
int Endx,endy;
int mapw,maph;
int steps;
int xopt[5]= {0,0,1,0,-1};
int yopt[5]= {0,1,0,-1,0};
int map[100][100]= {};
  void init () {x = 1;
  y = 1;
  EndX = 1;
  Endy = 9;
  Maph = 10;
  MAPW = 10; for (int i=0; i<=maph; i++) for (int j=0; j<=mapw; J + +) {if (i==0 | | j==0 | | i==maph| |
    J==MAPW) Map[i][j]=-1;
  } steps=0;
  Map[1][2]=-1;
  Map[2][2]=-1;
  Map[3][2]=-1;
 
  Map[4][2]=-1;
  Map[6][2]=-1;
  Map[7][2]=-1;
  Map[8][2]=-1;
  Map[9][2]=-1;
  Map[9][3]=-1;
  Map[8][3]=-1;
  Map[1][4]=-1;
  Map[3][4]=-1;
  Map[4][4]=-1;
  Map[5][4]=-1;
  Map[6][4]=-1;
  Map[7][4]=-1;
  Map[1][6]=-1;
  Map[2][6]=-1;
  Map[3][6]=-1;
  Map[4][6]=-1;
  Map[5][6]=-1;
  Map[6][6]=-1;
  Map[7][6]=-1;
  Map[8][6]=-1;
  Map[8][7]=-1;
  Map[8][8]=-1;
  Map[7][8]=-1;
  Map[6][8]=-1; Map[5][8]=-1;
  Map[4][8]=-1;
  Map[3][8]=-1;
  Map[2][8]=-1;
 
  Map[1][8]=-1;
map[endx][endy]=5;
  } void Dis () {System ("CLS");
  int ori = Map[x][y];
  Map[x][y]=1;
      for (int i=0; i<=maph; ++i) {for (int j=0; j<=mapw; ++j) {if (map[i][j]==0) printf ("");
      else if (map[i][j]==-1) printf ("#");
      else if (map[i][j]==1) printf ("@");
      else if (map[i][j]==2) printf (".");
    else if (map[i][j]==5) printf ("!");
  } cout<<i<<endl;
  for (int j=0; j<=mapw; ++j) cout<<j<< "";
  printf ("\ n > steps:%d Exit: (%d,%d) \ n", Steps,endx,endy);
Map[x][y] = the Ori;
  int can (int n) {if (map[x+xopt[n]][y+yopt[n]] = = 0 | | map[x+xopt[n]][y+yopt[n]] = = 5) return 1;
 
return 0;
  } void visit (int n) {map[x][y]=2;
  X+=xopt[n];
  Y+=yopt[n];
  node tem;
  tem.x = x;
  Tem.y = y;
  tem.lastopt = n;
  Sta.push_back (TEM);
steps++;
  int main () {init ();
  node tem; while (x!= endx | | y!=endy) {int cans = 0;
        for (int i=1; i<=4; i++) {if (Can (i)) {cans = 1;
        Visit (i);
      Break
        } if (!cans) {if (!sta.empty ()) {tem = Sta.back ();
        map[tem.x][tem.y]=0;
      Sta.pop_back ();
        else {map[x][y]=2;
        X+=XOPT[TEM.LASTOPT];
        X+=YOPT[TEM.LASTOPT];
        Dis ();
      Break
    } dis ();
 
  Sleep (500);
  } if (x==endx && y = = Endy) cout<< "\ n > I am finished....\n";
  else cout<< "\ n > I am finished...but I can ' t find the right way\n";
return 0; }

Effect Chart:

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