B - Dungeon Master-优快云博客

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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!

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>

using namespace std;

struct node
{
    int x;
    int y;
    int z;
    int num;
} link[40000];

int op, tp, l, m, n, flag, ans;
int v[44][44][44];///标记当前结点是否入队过
char st[44][44][44];///记录原始输入数据
int jk[6][3] = {{0, 1, 0}, {1, 0, 0}, {0, -1, 0}, {-1, 0, 0}, {0, 0, -1}, {0, 0, 1}};
///jk数组实现6个移动方向

void BFS();
bool Judge(int dz, int dx, int dy);

int main()
{
    int i, j, k;
    while(scanf("%d %d %d", &l, &m, &n) && (l != 0 || n != 0 || m != 0))
    {
        ///getchar();///吸收回车符
        op = tp = flag = 0;
        memset(v, 0, sizeof(v));
        for(i = 0; i < 40000; i++)
            link[i].num = 0;
        for(i = 0; i < l; i++)
        {
            for(j = 0; j < m; j++)
            {
                scanf("%s", st[i][j]);
                ///getchar();///吸收回车符
            }
            ///getchar();///吸收回车符
        }
        for(i = 0; i < l; i++)
        {
            for(j = 0; j < m; j++)
            {
                for(k = 0; k < n; k++)
                {
                    if(st[i][j][k] == 'S')
                    {
                        link[tp].x = j;///入队操作
                        link[tp].y = k;///入队操作
                        link[tp].z = i;///入队操作
                        tp++;///入队操作
                        v[i][j][k] = 1;///标记当前结点已入队
                        break;
                    }
                }
                if(k != n)
                    break;
            }
            if(j != m)
                break;
        }

        BFS();///广度优先搜索
        if(flag)
            printf("Escaped in %d minute(s).\n", ans);
        else
            printf("Trapped!\n");
    }
    return 0;
}
bool Judge(int dz, int dx, int dy)///判断移动方向结点是否合法
{
    if(dx < 0 || dx >= m || dy < 0 || dy >= n || dz < 0 || dz >= l)
        return false;
    else
        return true;
}
void BFS()///广度优先搜索
{
    int x, y, z, i;///注意不要混淆x,y,z的含义
    while(op < tp)
    {
        for(i = 0; i < 6; i++)
        {
            x = link[op].x + jk[i][0];
            y = link[op].y + jk[i][1];
            z = link[op].z + jk[i][2];
            if(Judge(z, x, y) && v[z][x][y] == 0)///判断移动方向结点是否符合合法和移动方向结点未入队过的条件
            {

                if(st[z][x][y] == '.')///判断移动方向结点是否符合可入队条件
                {
                    v[z][x][y] = 1;///标记移动方向结点入队
                    link[tp].x = x;
                    link[tp].y = y;
                    link[tp].z = z;
                    link[tp].num = link[op].num + 1;
                    tp++;
                }
                else if(st[z][x][y] == 'E')///判断是否到达目标结点
                {
                    v[z][x][y] = 1;///标记移动方向结点入队
                    flag = 1;
                    ans = link[op].num + 1;
                    break;
                }
            }
        }
        if(flag)///可到达目标结点
            break;
        op++;///当前结点出队
    }
}