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POJ - 2251 Dungeon Master

2018年07月30日  | 移动技术网IT编程  | 我要评论

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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!

大致题意：给定一个三维的迷宫，求出到达终点的最短时间

表面很复杂但实际上只是三维的BFS就不多说了直接看代码吧

 1 #include<stdio.h>
 2 #include<iostream>
 3 #include<string.h>
 4 #include<queue>
 5 using namespace std;
 6 struct node{
 7     int x,y,z,t;
 8 };
 9 int l,r,c,sx,sy,sz,tx,ty,tz,ans;
10 bool map[35][35][35],vis[35][35][35];
11 int dir[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};
12 bool check(int x,int y,int z)
13 {
14     if(x<0||y<0||z<0||x>r||y>c||z>l||map[x][y][z]==0||vis[x][y][z]) return 1;
15     return 0;
16 }
17 int bfs()
18 {
19     node now,next;
20     queue<node>q;
21     vis[sx][sy][sz]=1;
22     now.x=sx,now.y=sy,now.z=sz,now.t=0;
23     q.push(now);
24     while(!q.empty())
25     {
26         now=q.front();
27         q.pop();
28         if(now.x==tx&&now.y==ty&&now.z==tz) return now.t;
29         for(int i=0;i<6;++i)
30         {
31             next.x=now.x+dir[i][0];
32             next.y=now.y+dir[i][1];
33             next.z=now.z+dir[i][2];
34             if(check(next.x,next.y,next.z)) continue;
35             vis[next.x][next.y][next.z]=1;
36             next.t=now.t+1;
37             q.push(next);
38         }
39     }
40     return 0;
41 }
42 int main()
43 {
44     char s[35];
45     while(1)
46     {
47         memset(map,0,sizeof(map));
48         memset(vis,0,sizeof(vis));
49         scanf("%d%d%d",&l,&r,&c);
50         if(l==0&&r==0&&c==0) return 0;
51         for(int i=1;i<=l;++i)
52             for(int j=1;j<=r;++j)
53             {
54                 scanf("%s",s+1);
55                 for(int g=1;g<=strlen(s);++g)
56                 {
57                     if(s[g]=='S') sx=j,sy=g,sz=i;
58                     else if(s[g]=='E') tx=j,ty=g,tz=i,map[j][g][i]=1;
59                     else if(s[g]=='.') map[j][g][i]=1;
60                     else map[j][g][i]=0;
61                 }
62             }
63         ans=bfs();
64         if(ans) printf("Escaped in %d minute(s).\n",ans);
65         else printf("Trapped!\n");
66     }
67 }

 

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