Dungeon Master（三维bfs）

题目链接：http://poj.org/problem?id=2251

题目：

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).

L is the number of levels making up the dungeon.

R and C are the number of rows and columns making up the plan of each level.

Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.

If it is not possible to escape, print the line

Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

题意：你处在一个三维的地牢里，从S出发逃到出口E，问最少要跑多远。

思路：这题虽然是一个三维的地图，但是做法和二维的没多大区别，不过要从当前层到其他层的要求是你所在位置为非#，且你将到的那层的这个位置也是非#。

代码实现如下：

 #include <queue>
 #include <cstdio>
 #include <cstring>
 using namespace std;

 const int inf = 0x3f3f3f3f;
 int l, r, c, ans;
 int sx, sy, sz;
 char mp[][][];
 int vis[][][];

 struct node {
     int x, y, z, step;
 }nw, nxt;

 int dx[] = {, -, , , , }, dy[] = {, , , -, , },
     dz[] = {, , , , , -};

 void bfs(int z, int x, int y) {
     vis[z][x][y] = ;
     nw.z = z, nw.x = x, nw.y = y, nw.step = ;
     queue<node> q;
     q.push(nw);
     while(!q.empty()) {
         nw = q.front(), q.pop();
         if(mp[nw.z][nw.x][nw.y] == 'E') {
             ans = nw.step;
             return;
         }
         for(int i = ; i < ; i++) {
             nxt.z = nw.z + dz[i];
             nxt.x = nw.x + dx[i];
             nxt.y = nw.y + dy[i];
             if(nxt.z >=  && nxt.z < l && nxt.x >=  && nxt.x < r && nxt.y >= && nxt.y < c && vis[nxt.z][nxt.x][nxt.y] ==  && mp[nxt.z][nxt.x][nxt.y] != '#') {
                 nxt.step = nw.step + ;
                 vis[nxt.z][nxt.x][nxt.y] = ;
                 q.push(nxt);
             }
         }
     }
 }

 int main() {
     while(~scanf("%d%d%d", &l, &r, &c) && (l + r + c)) {
         for(int i = ; i < l; i++) {
             for(int j = ; j < r; j++) {
                 scanf("%s", mp[i][j]);
                 for(int k = ; k < c; k++) {
                     if(mp[i][j][k] == 'S') {
                         sx = j, sy = k, sz =i;
                     }
                 }
             }
         }
         memset(vis, , sizeof(vis));
         ans = inf;
         bfs(sz, sx, sy);
         if(ans >= inf) {
             printf("Trapped!\n");
         } else {
             printf("Escaped in %d minute(s).\n", ans);
         }
     }
     return ;
 }