See Data structure Write code (13) Application of Stack (IV.) Maze Solver

This is an example of solving a maze problem with a poor lifting method, but it is somewhat inadequate in the shortest path of efficiency and reconciliation .

These two questions remain in the next idle moment to answer.

Please note that the code is insufficient

The following code:

Maze.cpp: Defines the entry point of the console application. #include "stdafx.h" #include <stdlib.h>struct postion{int x;int y;}; typedef int DIRECTION;STRUCT elementtype{postion pos;direction Direction;}; Enum E_state{e_state_error = 0,E_STATE_OK = 1,};//Linked list node nodestruct lstacknode{elementtype Data;lstacknode * next;};/ /link Stack struct linkstack{lstacknode * bottom;lstacknode * top;int Len;}; Lstacknode * Makenode (ElementType data) {Lstacknode * Pnode = (Lstacknode *) malloc (sizeof (Lstacknode)); if (pnode! = NULL) { Pnode->data = Data;pnode->next = NULL;} return pnode;} E_state stackinit (Linkstack * lstack) {//Assign head node Lstacknode * Pnode = (Lstacknode *) malloc (sizeof (Lstacknode)); if (Pnode = = N ULL) {return e_state_error;} Pnode->next = null;//The top of the stack and the bottom of the stack are empty when they point to the same node. Lstack->bottom = Lstack->top = Pnode;lstack->len = 0;return E_State_Ok ;} void Stackclear (Linkstack * stack) {Lstacknode * next = Stack->bottom->next;while (next! = NULL) {Lstacknode * Freenod E = next;/* and careless ... Free (freenode); next = Next->next;*/next = next->next;free (freenode);} Stack->top = stack->bottom;//forgot//Error Stack->bottom->next = Null;stack->len = 0;} void Stackdestory (Linkstack * stack) {stackclear (stack); free (stack->top); stack->top = Stack->bottom = NULL;} E_state stackgettop (linkstack stack,elementtype * topdata) {//List of the stack top points to the top element if (stack.top! = stack.bottom) {*topdata = stack . Top->data;return E_state_ok;} Else{return E_state_error;}} int Stacklen (Linkstack stack) {return stack.len;} BOOL Stackempty (Linkstack stack) {return stack.top = = Stack.bottom? True:false;} E_state Stackpush (Linkstack * stack,elementtype data) {Lstacknode * node = makenode (data); if (node! = NULL) {stack->top- >next = Node;stack->top = node;stack->len++;} Else{return E_state_error;}} E_state Stackpop (linkstack * stack,elementtype * data) {if (stack->top! = Stack->bottom) {//First point to first element. Lstacknode * Next = Stack->bottom;*data = stack->top->data;//finds the precursor of the top element of the stack while (next->next! = stack->top) {next = next- >next;} FreE (stack->top); next->next = Null;stack->top = next;//forgot to add Stack->len--;return e_state_ok;} Else{return E_state_error;}} int mazedata[10][10] = {{0,0,0,0,0,0,0,0,0,0},{0,1,1,0,1,1,1,0,1,0},{0,1,1,0,1,1,1,0,1,0},{0,1,1,1,1,0,0,1,1,0},{ 0,1,0,0,0,1,1,1,1,0},{0,1,1,1,0,1,1,1,1,0},{0,1,0,1,1,1,0,1,1,0},{0,1,0,0,0,1,0,0,1,0},{0,0,1,1,1,1,1,1,1,0},{ Whether a node of the 0,0,0,0,0,0,0,0,0,0},};//pos coordinate can be passed bool Ispass (postion POS) {int pass = mazedata[pos.x][pos.y];//node is not a barrier, and Is the node that did not pass return pass = = 1? True:false;} Gets the next coordinate postion getnextpos (postion pos,direction di) {int addarray[][2] = {{0,1},{1,0},{0,-1},{-1,0}}; Postion nextpos;nextpos.x = pos.x + addarray[di][0];nextpos.y = pos.y + Addarray[di][1];return NextPos;} void Footpos (postion pos,int curstep) {Mazedata[pos.x][pos.y] = curstep;} void Delpos (postion pos) {Mazedata[pos.x][pos.y] =-1;} int Mazepath (postion startpos,postion endpos) {linkstack stack; ElementType data;int curstep = 2;stackinit (&stack);d o{if (Ispass (startpos)) {Footpos (Startpos,curstEP); curstep++;d ata.pos.x = startpos.x;data.pos.y = Startpos.y;data.direction = 0;stackpush (&stack,data); if ( Startpos.x = = Endpos.x && Startpos.y = = endpos.y) {stackdestory (&stack); return 1;} Else{startpos = Getnextpos (startpos,0);}} Else{if (!stackempty (Stack)) {Stackpop (&stack,&data); Curstep--;while (data.direction = = 3 &&!) Stackempty (Stack)) {//error forgot to delete: Delpos (Data.pos); Stackpop (&stack,&data); Curstep--;//startpos.x = Data.pos.x;//startpos.y = Data.pos.y;} if (Data.direction < 3) {data.direction++;//change direction to explore Stackpush (&stack,data); curstep++;startpos = Getnextpos ( data.pos,data.direction);}}} while (!stackempty (stack)); stackdestory (&stack); return 0;} Output solution void Printmaze () {int I,j;int x = 10,y = 10;for (i=0;i<x;i++) {for (j=0;j<y;j++) printf ("%3d", Mazedata[i][j]); printf ("\ n");} printf ("\ n");} int _tmain (int argc, _tchar* argv[]) {printf ("Initial maze as follows: \ n"); Printmaze (); Postion startpos; Postion endpos;printf ("Please enter the starting position coordinates x, Y (from 0 rows, 0 columns):"); scanf ("%d%d", &startpos.x,&startpos.y);p rintf ("Please enter the terminating position coordinates x, Y (from 0 rows, 0 columns):"); scanf ("%d%d", &endpos.x,&endpos.y); int result = Mazepath (Startpos,endpos), if (result = = 1) {printf ("Maze solution is as follows: (start address 2) \ n"); Printmaze ();} Else{printf ("Maze from (%d,%d) to (%d,%d) no solution \ n", startpos.x,startpos.y,endpos.x,endpos.y); Printmaze ();} return 0;}

The initial maze of 0 represents the wall, and 1 represents the pathway.

See Data structure Write code (13) Application of Stack (IV.) Maze Solver