poj2251(bfs寻找最短路径，三位迷宫)

Posted 2021-01-18 codeoosacm

题目链接：http://poj.org/problem?id=2251

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 50755		Accepted: 19053

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!

Source

Ulm Local 1997

比较简单，多输入几层迷宫，4个方向变成6个方向就ok了.（我也不知道为什么，实验课偷偷做题被这道sb题卡半天。。。可能是报复吧QAQ）

需要注意的坑点是，

输入顺序，先是z轴，再是xy的二维平面

代码附上：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 #include<string>
 5 #include<cstring>
 6 #define sca(a) scanf("%d",&a );
 7 #define sca2(a,b) scanf("%d%d",&a,&b );
 8 #define sca3(a,b,c) scanf("%d%d%d",&a,&b,&c);
 9 #define mem(a,b) memset(a,b,sizeof(a));
10 #define FOR0(i,j) for(int i=0;i<=j;++i)
11 #define FOR1(i,j) for(int i=1;i<=j;++i)
12 using namespace std;
13 struct node
14 {
15     int x,y,z,step;
16 };
17 char plat[40][40][40];
18 int vis[40][40][40];
19 int dir[6][3]={0,0,1,0,1,0,1,0,0,-1,0,0,0,-1,0,0,0,-1};
20 int m,n,o;
21 int sx,sy,sz;
22 void ini()
23 {
24     mem(plat,0);
25     mem(vis,0);
26 }
27 int main()
28 {
29     //cout<<sx<<sy<<sz;
30     while(scanf("%d%d%d",&o,&m,&n)&&m+n+o)
31     {
32         ini();
33         for(int i=1;i<=o;i++)
34         {
35             for(int j=1;j<=m;j++)
36             {
37                 scanf("%s", plat[i][j]+1);
38                 for(int k=1;k<=n;k++)
39                 {
40                     if(plat[i][j][k]==‘S‘)
41                     {
42                         sx=j;sy=k;sz=i;
43                     }
44                 }
45             }
46         }
47         //cout<<sx<<sy<<sz;
48         queue<node>q;
49         q.push({sx,sy,sz,0});
50         vis[sz][sx][sy]=1;
51         int flag=0;
52         while(!q.empty())
53         {
54             node now=q.front();
55             q.pop();
56             if(plat[now.z][now.x][now.y]==‘E‘)
57             {
58                 flag=1;
59                 printf("Escaped in %d minute(s).
", now.step);
60                 break;
61             }
62             int nex,ney,nez;
63             for(int i=0;i<6;i++)
64             {
65                 nex=now.x+dir[i][0];
66                 ney=now.y+dir[i][1];
67                 nez=now.z+dir[i][2];
68                 if(nex<1||nex>m||ney<1||ney>n||nez<1||nez>o) continue;
69                 if(vis[nez][nex][ney]) continue;
70                 if(plat[nez][nex][ney]==‘.‘||plat[nez][nex][ney]==‘E‘)
71                 {
72                     //printf("%d %d %d
",nex,ney,nez );
73                     vis[nez][nex][ney]=1;
74                     q.push({nex,ney,nez,now.step+1});
75                 }
76             }
77         }
78         if(!flag)
79             printf("Trapped!
");
80 }
81 return 0;
82 
83 }