2-5-6264: Out of the maze

Describe

When you are standing in a maze, you tend to lose your sense of direction by the intricacies of the road, and if you can get a maze map, things can become very simple.
Assuming you've got a blueprint for a n*m maze, please find the shortest path from the starting point to the exit.

The first line of input is two integers n and m (1<=n,m<=100), representing the number of rows and columns of the maze.
Next n rows, each line of a string of length m, represents the layout of the entire maze. Character '. ' Represents the open space, ' # ' represents the wall, ' S ' represents the starting point, ' T ' represents the exit. The minimum number of steps to take to output output from the starting point to the exit. Sample input

3 3s#t.# ....

Sample output

6

1#include <stdio.h>2#include <string.h>3 intni[4]={0,1,0,-1};4 intnj[4]={1,0,-1,0};5 intqu[100 on][2],dep[100xx];6 Chara[101][101];7 intBook[101][101];8 intm,n,i,j,k,si,sj,xi,xj,qi,qj,zi,zj,bu=0, ans;9 intMain ()Ten { One     inth,t; Ascanf"%d%d",&m,&n); -memset (book,0,sizeof(book)); -      for(i=1; i<=m;i++) thescanf"%s", A[i]); -      for(i=1; i<=m;i++) -          for(j=0; j<=n-1; j + +) -         { +             if(a[i][j]=='S') {qi=i;qj=j;book[i][j]=1;} -             if(a[i][j]=='T') {zi=i;zj=J;} +         } AH=0; t=0; ans=-1; atmemset (DEP,0,sizeof(DEP)); -Memset (Qu,0,sizeof(qu)); -qu[0][0]=qi;qu[0][1]=QJ; -          while(h<=t) -         { -xi=qu[h][0]; inxj=qu[h][1]; -             if(xi==zi&&xj==ZJ) to             { +ans=Dep[h]; -                  Break; the             } *              for(k=0; k<=3; k++) $             {Panax Notoginsengsi=xi+Ni[k]; -sj=xj+Nj[k]; the                 if(si>=1&&si<=m&&sj>=0&&sj<=n-1) +                     if(a[si][sj]!='#'&&book[si][sj]!=1) A                     { thequ[++t][0]=si; +qu[t][1]=SJ; -dep[t]=dep[h]+1; $book[si][sj]=1; $                     } -             } -h++; the         } -printf"%d\n", ans);Wuyi     return 0; the}

2-5-6264: Out of the maze