Dungeon Master
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 10668 |
|
Accepted: 4111 |
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!
這題不難，廣搜，不知道為什么這個題目在那個指導中出現(xiàn)在dfs中
代碼也好些，就是floodfill，往六個方向搜，如果已經(jīng)走過則不走，如果沒走過且能走則走
這次數(shù)組開小了，第一次隊列開了1000，不知道怎么想的
第二次檢查 改成10000了，然后還是wa
第三次 改成100000，終于過了
1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4struct node
5
{
6
int x,y,z,d;
7
};
8int dx[6][3]= {{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
9int l,r,c,sum,ans,head,tail;
10struct node que[100000],ss,tt;
11short hash[35][35][35];
12char map[35][35][35];
13void bfs()
14{
15 int i,j;
16 struct node now,new1;
17 ans=0;
18 head=0;
19 tail=1;
20 que[tail]=ss;
21 memset(hash,0,sizeof(hash));
22 hash[ss.x][ss.y][ss.z]=1;
23 while (head<tail)
24
{
25 head++;
26 now=que[head];
27 /**//*if (now.x==tt.x&&now.y==tt.y&&now.z==tt.z)
28 {
29 ans=now.d;
30 return;
31
}*/
32 for (i=0; i<6 ; i++ )
33 {
34 new1.x=now.x+dx[i][0];
35 new1.y=now.y+dx[i][1];
36 new1.z=now.z+dx[i][2];
37 new1.d=now.d+1;
38 if (new1.x>=0&&new1.x<l&&new1.y>=0&&new1.y<r&&new1.z>=0&&new1.z<c)//未越界
39 if ((map[new1.x][new1.y][new1.z]!='#')&&(!hash[new1.x][new1.y][new1.z]))//能走且未走過
40 {
41 if (map[new1.x][new1.y][new1.z]=='E')
42 {
43 ans=new1.d;
44 return;
45 }
46 if (!hash[new1.x][new1.y][new1.z])
47 {
48 tail++;
49 que[tail]=new1;
50 hash[new1.x][new1.y][new1.z]=1;
51 }
52 }
53 }
54 }
55}
56void init()
57{
58 int i,j,k;
59 for (i=0; i<l; i++ )
60 {
61 for (j=0; j<r ; j++ )
62 {
63 scanf("%s",&map[i][j]);
64 for (k=0; k<c ; k++ )
65 {
66 if (map[i][j][k]=='S')
67 {
68 ss.x=i;
69 ss.y=j;
70 ss.z=k;
71 ss.d=0;
72 }
73 if (map[i][j][k]=='E')
74 {
75 tt.x=i;
76 tt.y=j;
77 tt.z=k;
78 }
79 }
80 }
81 }
82}
83int main()
84{
85 while (scanf("%d%d%d",&l,&r,&c)!=EOF&&!(l==0&&r==0&&c==0))
86 {
87 init();
88 bfs();
89 if (ans==0)
90 {
91 printf("Trapped!\n");
92 }
93 else
94 printf("Escaped in %d minute(s).\n",ans);
95 }
96 return 0;
97}
98