Dungeon Master
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6561 |
|
Accepted: 2610 |
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!
三維迷宮問題
據(jù)說要用BFS，范圍30 ，我以為DFS能搞定的，為什么會超時呢？？？？
1
#include<iostream>
2
#include<string.h>
3
using namespace std;
4
char G[35][35][35];
5
bool g[35][35][35];
6
void dfs(int l,int r,int c)
7
{
8
if(G[l][r][c]==0||G[l][r][c]=='#')return ;
9
if((G[l+1][r][c]=='.'||G[l+1][r][c]>G[l][r][c])&&G[l+1][r][c]!=0&&G[l+1][r][c]!='#')
{ G[l+1][r][c]=G[l][r][c]+1;g[l+1][r][c]=1;dfs(l+1,r,c);}
10
if((G[l][r-1][c]=='.'||G[l][r-1][c]>G[l][r][c])&&G[l][r-1][c]!=0&&G[l][r-1][c]!='#')
{ G[l][r-1][c]=G[l][r][c]+1;g[l][r-1][c]=1;dfs(l,r-1,c);}
11
if((G[l][r+1][c]=='.'||G[l][r+1][c]>G[l][r][c])&&G[l][r+1][c]!=0&&G[l][r+1][c]!='#')
{ G[l][r+1][c]=G[l][r][c]+1;g[l][r+1][c]=1;dfs(l,r+1,c);}
12
if((G[l][r][c-1]=='.'||G[l][r][c-1]>G[l][r][c])&&G[l][r][c-1]!=0&&G[l][r][c-1]!='#')
{ G[l][r][c-1]=G[l][r][c]+1;g[l][r][c-1]=1;dfs(l,r,c-1);}
13
if((G[l][r][c+1]=='.'||G[l][r][c+1]>G[l][r][c])&&G[l][r][c+1]!=0&&G[l][r][c+1]!='#')
{ G[l][r][c+1]=G[l][r][c]+1;g[l][r][c+1]=1;dfs(l,r,c+1);}
14
if((G[l-1][r][c]=='.'||G[l-1][r][c]>G[l][r][c])&&G[l-1][r][c]!=0&&G[l-1][r][c]!='#')
{ G[l-1][r][c]=G[l][r][c]+1;g[l-1][r][c]=1;dfs(l-1,r,c);}
15
}
16
17
int main()
18
{
19
int i,j,k,l,r,c,sl,sr,sc,el,er,ec; //起點，終點
20
while(cin>>l>>r>>c,l||r||c)
21
{
22
memset(G,0,sizeof (G));
23
memset(g,0,sizeof (g));
24
for(i=1; i<=l ;i++)
25
for(j=1; j<=r; j++)
26
for(k=1; k<=c; k++)
27
{
28
cin>>G[i][j][k];
29
if(G[i][j][k]=='S')
30
{ sl=i; sr=j; sc=j; G[i][j][k]=1; g[i][j][k]=1; }
31
else if(G[i][j][k]=='E')
32
{el=i; er=j; ec=k; G[i][j][k]=127; }
33
}
34
35
dfs(sl,sr,sc);
36
/**//* for(i=1; i<=l ; i++,cout<<endl)
37
for(j=1; j<=r; j++,cout<<endl)
38
for(k=1; k<=c; k++)
39
cout<<(int)G[i][j][k]<<" ";
40
for(i=1; i<=l ; i++,cout<<endl)
41
for(j=1; j<=r; j++,cout<<endl)
42
for(k=1; k<=c; k++)
43
cout<<(int)g[i][j][k]<<" ";
44
*/if(g[el][er][ec]==0)cout<<"Trapped!"<<endl;
45
else cout<<"Escaped in "<<int(G[el][er][ec]-1)<<" minute(s)."<<endl;
46
}
47
// system("pause");
48
return 0;
49
}
寫個BFS終于過了，只要在平面的迷宮基礎(chǔ)上在垂直方向往上，往下都搜兩次就可以了
1 #include<iostream>
2 #include<string.h>
3 #include<queue>
4 using namespace std;
5 char G[35][35][35];
6 int cnt[35][35][35];
7 struct type
8 { int l,r,c; };
9
10 int main()
11 {
12 int l,c,r,sl,sc,sr,el,ec,er,i,j,k;
13 void bfs(int sl,int sr,int sc);
14 while(cin>>l>>r>>c,l||c||r)
15 {
16 memset(G,0,sizeof (G));
17 memset(cnt,0,sizeof (cnt));
18 for(i=1; i<=l; i++)
19 for(j=1; j<=r; j++)
20 for(k=1; k<=c; k++)
21 {
22 cin>>G[i][j][k];
23 if(G[i][j][k]=='S'){sl=i; sr=j; sc=k; }
24 else if(G[i][j][k]=='E'){el=i; er=j; ec=k; }
25 }
26
27 bfs(sl,sr,sc);
28 // for(i=1; i<=l; i++,cout<<endl)
29 // for(j=1; j<=r; j++,cout<<endl)
30 //for(k=1; k<=c; k++)
31 //cout<<cnt[i][j][k]<<' ';
32
33 if(el==sl&&er==sr&&ec==sc)cout<<"Escaped in "<<0<<" minute(s). "<<endl;
34 else if(cnt[el][er][ec]>0)cout<<"Escaped in "<<cnt[el][er][ec]<<" minute(s). "<<endl;
35 else cout<<"Trapped! "<<endl;
36 }
37 // system("pause");
38 return 0;
39 }
40
41 bool valid(int l,int r,int c)
42 {
43 return (G[l][r][c]=='.'||G[l][r][c]=='E');
44 }
45
46 void bfs(int sl,int sr, int sc)
47 {
48 queue<struct type> q;
49 struct type tem; tem.l=sl; tem.r=sr; tem.c=sc;
50 q.push(tem);
51 while(q.size())
52 {
53 tem=q.front(); q.pop();
54 int l=tem.l,r=tem.r,c=tem.c;
55 if(valid(l,r+1,c)&&cnt[l][r+1][c]==0){cnt[l][r+1][c]=cnt[l][r][c]+1;tem.l=l; tem.r=r+1; tem.c=c; q.push(tem); }
56 if(valid(l,r-1,c)&&cnt[l][r-1][c]==0){cnt[l][r-1][c]=cnt[l][r][c]+1;tem.l=l; tem.r=r-1; tem.c=c; q.push(tem); }
57 if(valid(l,r,c+1)&&cnt[l][r][c+1]==0){cnt[l][r][c+1]=cnt[l][r][c]+1;tem.l=l; tem.r=r; tem.c=c+1; q.push(tem); }
58 if(valid(l,r,c-1)&&cnt[l][r][c-1]==0){cnt[l][r][c-1]=cnt[l][r][c]+1;tem.l=l; tem.r=r; tem.c=c-1; q.push(tem); }
59 if(valid(l-1,r,c)&&cnt[l-1][r][c]==0){cnt[l-1][r][c]=cnt[l][r][c]+1;tem.l=l-1; tem.r=r; tem.c=c; q.push(tem); }
60 if(valid(l+1,r,c)&&cnt[l+1][r][c]==0){cnt[l+1][r][c]=cnt[l][r][c]+1;tem.l=l+1; tem.r=r; tem.c=c; q.push(tem); }
61
62 }
63 }