Dungeon Master

给定一个多层的地牢（多维数组），'.'表示空白格， '#'表示障碍。给定一个起点和终点，可以在空白格移动，可以向上移动，向下移动，向左移动和向右移动，问到达终点的最短路径。

思路：BFS

#include<iostream>
#include<queue>

using namespace std;

typedef struct pos{
  int x;
  int y;
  int z;
}Pos;

const int INF = 1000000;
int L, R, C;
int sx, sy, sz;
int fx, fy, fz;
char maze[40][40][40];
int d[40][40][40];
int dx[6] = {1, -1, 0, 0, 0, 0}, dy[6] = {0, 0, 0, 0, 1, -1}, dz[6] = {0, 0, -1, 1, 0, 0};

void bfs(int x, int y, int z){
  Pos p;
  p.x = x;
  p.y = y;
  p.z = z;
  queue<Pos> q;
  q.push(p);
  d[x][y][z] = 0;

  while( !q.empty() ){
    Pos t = q.front(); q.pop();

    for(int i=0; i<6; i++){
      int nx = t.x + dx[i];
      int ny = t.y + dy[i];
      int nz = t.z + dz[i];

      if(nx >= 0 && nx < L && ny >= 0 && ny < R && nz >= 0 && nz < C && d[nx][ny][nz] == INF && maze[nx][ny][nz] != '#'){
        Pos rel;
        rel.x = nx;
        rel.y = ny;
        rel.z = nz;
        q.push(rel);
        d[nx][ny][nz] = d[t.x][t.y][t.z] + 1;
      }
    }

  }
}

int main(){
  ios::sync_with_stdio(false);

  while(cin >> L >> R >> C){
    if(L==0 && R==0 && C==0)
      break;

    for(int i=0; i<L; i++){
      for(int j=0; j<R; j++){
        for(int k=0; k<C; k++){
          d[i][j][k] = INF;
          cin >> maze[i][j][k];
          if(maze[i][j][k] == 'S'){
            sx = i;
            sy = j;
            sz = k;
          }
          if(maze[i][j][k] == 'E'){
            fx = i;
            fy = j;
            fz = k;
          }
        }
      }
    }

    bfs(sx, sy, sz);
    if(d[fx][fy][fz] != INF)
      cout << "Escaped in " << d[fx][fy][fz] << " minute(s)." << endl;
    else
      cout << "Trapped!" << endl;
  }


  return 0;
}