Dungeon Master （广搜）

• 问题描述：

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!

• 解题思路：

用三维数组来存地图，然后用广度优先搜索即可。

• 代码：

#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
using namespace std;
char a[35][35][35];
int l,r,c;
int next[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
struct node
{
    int x;
    int y;
    int z;
    int step;
}s,e,now,net;

int bfs()
{
    queue<node> q;
    q.push(s);
    while(q.size())
    {
        now=q.front();
        if(now.x==e.x&&now.y==e.y&&now.z==e.z)
            return now.step;
        for(int i=0;i<6;i++)
        {
            net.x=now.x+next[i][0];
            net.y=now.y+next[i][1];
            net.z=now.z+next[i][2];
            if(net.x>0&&net.x<=l&&net.y>0&&net.y<=r&&net.z>0&&net.z<=c&&a[net.x][net.y][net.z]!='#')
            {
                a[net.x][net.y][net.z]='#';
                net.step=now.step+1;
                q.push(net);
            }
        }
        q.pop();
    }
    return -1;
}

int main()
{
    while(~scanf("%d%d%d",&l,&r,&c))
    {
        if(l==0&&r==0&&c==0)break;
        for(int i=1;i<=l;i++)
        {
            for(int j=1;j<=r;j++)
            {
                for(int k=1;k<=c;k++)
                {
                    scanf(" %c",&a[i][j][k]);
                    if(a[i][j][k]=='S')
                    {
                        s.x=i;
                        s.y=j;
                        s.z=k;
                        s.step=0;
                    }
                    if(a[i][j][k]=='E')
                    {
                        e.x=i;
                        e.y=j;
                        e.z=k;
                    }
                }
            }
        }
        int ans=bfs();
        if(ans==-1)printf("Trapped!\n");
        else printf("Escaped in %d minute(s).\n",ans);
    }
    return 0;
}