Maze Game console C++ code

This article Shared an example of C++ design of a resized maze game, given the entrance to the maze. If there is an exit, the program can display the path of walking, and finally reach the exit, and output "successfully out of the maze"; If there is no exit, the program can also show the progress of the walk, and finally retreat to the entrance, and output "retreat to the entrance."

// This is a 1 A maze game 
#include<iostream>
#include<ctime>
#include<cstdlib>/* Used to generate random Numbers to form a random maze */
#include<iomanip>/* Used for output format control to make the maze appearance rules generated */
using namespace std;
/* The function that generates the maze mazegenerator*/
char*mazegenerator(int m,int n)
{
 int i,j,k;/* counter */
 char *p=new char[m*n];/* With length of m*n Array store mxn A maze of size */
 srand(int(time(0)));/* Used to generate random Numbers */
 if(m<10||n<10)// Dense when the order is small 1 Dot, make it interesting 
 for(i=0;i<m*n;i++)
 {
 if(rand()%3==0)p[i]='.';// A random number for 3 Is defined as the path in the maze 
 else p[i]='#';
 }
 else
 for(i=0;i<m*n;i++)
 {
 if(rand()%2==0)p[i]='.';/* When the random number is even, the point in the maze is defined as the path */
 else p[i]='#';
 }
 /* Seal both the left and the top of the boundary except for one 1 Six entrances, only from the right and the bottom, 
  Make the maze more challenging */
 for(i=0;i<m;i++)
 p[i]='#';
 for(i=0;i<m*n;i+=m)
 p[i]='#';
 for(i=m-1;i<m*n;i+=m)
 p[i]='#';
 for(i=m*(n-1);i<m*n;i++)
 p[i]='#';
 /* Let's put the front 3 The order structure of T font */
 p[0]=p[1]=p[2]=p[2*m]=p[2*m+2]='#';
 p[m+1]=p[m+2]=p[2*m+1]='.';
 /* Maze entry is defined as 'x'*/
 p[m]='x';
 /* Mazes are random and need to be constructed 1 pathway */
 /* This is a 1 Each article 3 Turn, 1 The access of the time */
 for(j=3;j<m;j+=3)
 {
 for(k=0;k<3;k++)
 {
 i=(j-3)+k+1;
 if(i*m+j<m*n)
 p[i*m+j]='.';
 }
 }
 for(i=4;i<m;i+=3)
 for(k=0;k<3;k++)
 {
 j=(i-3)+k+2;
 if(i*m+j<m*n)
 p[i*m+j]='.';
 }
 
 return p;
}
/* You need to output a new maze after creating or modifying it */
void showmaze(char*p,int m,int n)
{
 int i,j;/*i On behalf of the line, j On behalf of the column, i*n+j Is the ordinal number of that point in the array */
 for(i=0;i<n;i++)
 {
 for(j=0;j<m;j++)
 cout<<setw(2)<<p[i*m+j];/* Each character takes two to square the maze */
 cout<<endl;
 }
}
/* Maze function mazetraverse*/
void mazetraverse(char*p,int m,int n)
{
 int x,y,z;/*y Is the current position, x Is the former 1 Step, z Is after 1 Step, t Used to enter the command to continue through the maze */
 /* Initialize the x,y, The location of the entrance to the maze */
 x=m;
 y=m+1;
 p[y]='x';/* This is the end of the maze 1 step */
 cout<<" Please press enter to move :";/* Let the user control the maze process */
 cin.get();
 cout<<endl;
 showmaze(p,m,n);/* Output the maze */
 cout<<endl;
 cout<<" Please press enter to move : ";
 cin.get();
 /* The point is to use x,y determine z The value of the */
 while(!((y>m*(n-1))||(((y+1)%m)==0)))/* Perform a loop through the maze before reaching the boundary */
 {
 /* Determine the direction of the right, if the right is '.', The right */
 if(y-x==1)z=y+m;
 if(y-x==m)z=y-1;
 if(y-x==-m)z=y+1;
 if(y-x==-1)z=y-m;
 /* The right not to '.' when y The front of the walk */
 if(p[z]=='#')
 {
 /* judge y The front of theta, if the front is theta '.', The forward */
 if(y-x==1)z=y+1;
 if(y-x==m)z=y+m;
 if(y-x==-m)z=y-m;
 if(y-x==-1)z=y-1;
 /* If the front is '#' Go to the left */
 if(p[z]=='#')
 {
 /* Determine the left position */
 if(y-x==1)z=y-m;
 if(y-x==m)z=y+1;
 if(y-x==-m)z=y-1;
 if(y-x==-1)z=y+m;
 /* If left to '#' To die we need to go backwards */
 if(p[z]=='#')z=x;
 }
 }
 p[z]='x';/* In the end, let it all go z The location of the for 'x', Represents the trace of a journey */
 showmaze(p,m,n);
 cout<<endl;
 cout<<" Please press enter to move : ";
 cin.get();
 cout<<endl;
 /* replace x,y Move the current position */
 x=y;
 y=z;
 }
}
int main()
{
 int m,n;/*n Is the order of the maze. Due to the limit of output window size, 
 n It cannot be more than the number of characters per line */
 cout<<"  Chen Weihang's maze game \n My maze is very neat! ^_^\n"<<endl;
 cout<<" This is a 1 A maze of arbitrary size ,"<<endl;
 cout<<" Each maze is different. \n"<<endl;
 cout<<" Because of the screen 1 Second most display 80 A character, \n Here, each character takes up two more Spaces, "<<endl
 <<" So the input exceeds 40 The steps are not pretty, \n It's still true. Ha ha \n"<<endl;
 cout<<" Please enter the width of the maze in turn m And highly n(m>2,n>2):"<<endl;
 /* You can test multiple sets of data */
 while(cin>>m>>n)
 {
 char*a=mazegenerator(m,n);
 showmaze(a,m,n);
 cout<<endl;
 cout<<" Long March 1 Step: \n"<<endl;
 mazetraverse(a,m,n);
 cout<<" Congratulations to you! You've made it through the maze! "<<endl<<endl;
 cout<<" To try a new maze, enter the order of the new maze n(n>2) . \n Otherwise, press the ctrl+z And then according to the enter The end of the "<<endl;
 }
}