Data structure Routines-maze problem (with stack structure)

This paper is aimed at the basic series of data Structure network course (3): Stack and queue of the 6th lesson stack Application 2-maze problem.

Example: Finding the path from the inlet to the exit

Program implementation:

#include <stdio.h> #define MAXSIZE #define M 8 #define N 8 int mg[m+2][n+2]= {{1,1,1,1,1,1,1,1,1,1}, {1 , 0,0,1,0,0,0,1,0,1}, {1,0,0,1,0,0,0,1,0,1}, {1,0,0,0,0,1,1,0,0,1}, {1,0,1,1,1,0,0,0,0,1}, {1,0,0,0,1,0,0,0
, 0,1}, {1,0,1,0,0,0,1,0,0,1}, {1,0,1,1,1,0,1,1,0,1}, {1,1,0,0,0,0,0,0,0,1}, {1,1,1,1,1,1,1,1,1,1}};              typedef struct {int i;              The line number of the current block Int J;             The column number of the current block int di;
Di is the next azimuth number of the adjacent position} Box;
    typedef struct {Box data[maxsize];            int top;               Stack top pointer} sttype;
    Defines the stack type int mgpath (int xi,int yi,int xe,int ye)//Solution path: (Xi,yi)--(Xe,ye) {int i,j,k,di,find;                  Sttype St;                  Define Stack St St.top=-1;                   Initializes the top pointer of the stack st.top++;
    Initial block into the stack st.data[st.top].i=xi;
    St.data[st.top].j=yi;
    St.data[st.top].di=-1;
    Mg[xi][yi]=-1;
        while (st.top>-1)//stack is not empty when looping {i=st.data[st.top].i; J=ST.DATA[ST.TOP].J;  Di=st.data[st.top].di;
            Take stack top box if (i==xe && j==ye)//Find exit, Output path {printf ("Maze path: \ n");
                for (k=0; k<=st.top; k++) {printf ("\ t (%d,%d)", ST.DATA[K].I,ST.DATA[K].J);
            if ((k+1)%5==0)//output every 5 blocks after a line of printf ("\ n");
            } printf ("\ n");      return (1);
        Returns 1} find=0 after finding a path;
            while (di<4 && find==0)//Find the next removable block {di++;
                Switch (DI) {case 0:i=st.data[st.top].i-1;
                J=ST.DATA[ST.TOP].J;
            Break
                Case 1:I=ST.DATA[ST.TOP].I;
                j=st.data[st.top].j+1;
            Break
                Case 2:i=st.data[st.top].i+1;
                J=ST.DATA[ST.TOP].J;
            Break Case 3:i=st.data[st.top].i,j=St.data[st.top].j-1;
            Break    } if (mg[i][j]==0) find=1;      Find Next adjacent block} if (find==1)//Find the next removable block {st.data[st.top].di=di;                   Modifies the DI value st.top++ of the top element of the original stack;
            Next can walk square into the stack st.data[st.top].i=i;
            St.data[st.top].j=j;
            St.data[st.top].di=-1;                Mg[i][j]=-1; Avoid repeatedly walking to the block} else//no path to go, then stack {mg[st.data[st.top].i][st.                   data[st.top].j]=0;//let the location become another path can walk block st.top--;                          Return The Block}} return (0);
    Indicates that there is no path to go, return 0} int main () {Mgpath (1,1,m,n);
return 0;
 }