Dungeon Master
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13923 | Accepted: 5424 |
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.
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!
这道题最坑人的地方在于poj上将他分类为dfs，结果一写就超时，其实写最短路一般肯定是bfs，哎，经验还是太浅了。
这道题还有一个值得注意的地方就是这是个3维的迷宫，只需要加上上下两个方向就行了，其他的就是简单的bfs。
这道题wa了半天，居然是我忘记把文件输入输出那句话给删了，尼玛！
下面是代码：
#include <cstdio>
#include <cstring>
#include <iostream>
#define pan(a,b,c) (a<=b&&b<=c)
using namespace std; int dir[6][3]={{0,0,-1},{0,0,1},{-1,0,0},{1,0,0},{0,-1,0},{0,1,0}};
char map[35][35][35];
int vis[35][35][35];
int t,m,n,flag;
struct node{
int x,y,z;
int num;
}que[30010]; void bfs(int sx,int sy,int sz){
int head=0;
int tail=1;
que[0].x=sx;
que[0].y=sy;
que[0].z=sz;
que[0].num=0;
vis[sx][sy][sz]=1;
flag=0;
int xx,yy,zz;
while(head<tail){
for(int i=0;i<6;i++){
xx=que[head].x+dir[i][0];
yy=que[head].y+dir[i][1];
zz=que[head].z+dir[i][2];
if(map[xx][yy][zz]=='E'){//搜到终点
printf("Escaped in %d minute(s).\n",que[head].num+1);
flag=1;
break;
}
if(pan(1,xx,t)&&pan(1,yy,m)&&pan(1,zz,n)&&vis[xx][yy][zz]==0&&map[xx][yy][zz]=='.'){//如果在迷宫内，并且未走过
que[tail].x=xx;
que[tail].y=yy;
que[tail].z=zz;
que[tail].num=que[head].num+1;
vis[xx][yy][zz]=1;
tail++;
}
}
if(flag)
break;
head++;//flag不为1，那就要在继续往后走
}
if(!flag)
cout<<"Trapped!"<<endl;
} int main(){
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
int i,j,k;
int sx,sy,sz;
while(scanf("%d%d%d",&t,&m,&n)!=EOF){
if(t==0&&m==0&&n==0) break;
for(i=1;i<=t;i++)
for(j=1;j<=m;j++)
for(k=1;k<=n;k++){
cin>>map[i][j][k];
if(map[i][j][k]=='S'){//找到起点
sx=i;
sy=j;
sz=k;
}
}
memset(vis,0,sizeof(vis));
bfs(sx,sy,sz);
}
return 0;
}