UVA 532 Solution

Problem Link
Problem Type : Single Source Shortest Path on Unweighted Graph (BFS).
Difficulty Level : Easy
Author : Imdad

Briefly problem statement
A 3d maze is given. We have to find out shortest path between a start point and a end point. Simple BFS can do the trick.
Here, is the C++ implementation of this problem.
 

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <queue>
#include <cstring>

using namespace std;

#define sz 31

int L,R,C;
char in[sz][sz][sz];
bool used[sz][sz][sz];

typedef struct{
    int L,R,C;
    int step;
}data;
data st,en,tmp,cur;

int solve(data start){
    queueQ;

 int dr[] = {0,0,-1,1};
 int dc[] = {-1,1,0,0};
 int i;

 memset(used,true,sizeof(used));
 Q.push(start);
 used[st.L][st.R][st.C] = false;

 while(!Q.empty()){
  cur = Q.front();
  Q.pop();
  if(in[cur.L][cur.R][cur.C] == 'E')
   return cur.step;
  //used[cur.L][cur.R][cur.C] = false;

  for(i = 0; i < 4; i++){
   tmp = cur;
   tmp.R += dr[i];
   tmp.C += dc[i];

   if(tmp.R < 0 || tmp.R >= R)
    continue;
   if(tmp.C < 0 || tmp.C >= C)
    continue;
   if(used[tmp.L][tmp.R][tmp.C] && in[tmp.L][tmp.R][tmp.C] != '#'){
    tmp.step = cur.step + 1;
    Q.push(tmp);
    used[tmp.L][tmp.R][tmp.C] = false;
   }
  }

  if(cur.L - 1 >= 0){
   tmp = cur;
   tmp.L--;
   if(used[tmp.L][tmp.R][tmp.C] && in[tmp.L][tmp.R][tmp.C] != '#'){
    tmp.step = cur.step + 1;
    Q.push(tmp);
    used[tmp.L][tmp.R][tmp.C] = false;
   }
  }

  if(cur.L + 1 < L){
   tmp = cur;
   tmp.L++;
   if(used[tmp.L][tmp.R][tmp.C] && in[tmp.L][tmp.R][tmp.C] != '#'){
    tmp.step = cur.step + 1;
    Q.push(tmp);
    used[tmp.L][tmp.R][tmp.C] = false;
   }
  }
 }
 return -1;
}
int main(){
 int i,j,k,ans;

 //freopen("i.txt","r",stdin);

 while(scanf("%d %d %d",&L,&R,&C) && !(!L && !R && !C) ){
     for(i = 0; i < L; i++){
         getchar();
         for(j = 0; j < R; j++){
             for(k = 0; k < C; k++){
                 in[i][j][k] = getchar();
                 if(in[i][j][k] == 'S'){
                     st.L = i;
                     st.R = j;
                     st.C = k;
                     st.step = 0;
                 }
                }
             getchar();
         }
     }
  ans = solve(st);
  if(ans == -1)
   printf("Trapped!\n");
  else
   printf("Escaped in %d minute(s).\n",ans);

 }


 return 0;
}