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.
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!這道題最坑人的地方在於poj上將他分類為dfs，結果一寫就超時，其實寫最短路一般肯定是bfs，哎，經驗還是太淺了。這道題還有一個值得注意的地方就是這是個3維的迷宮，只需要加上上下兩個方向就行了，其他的就是簡單的bfs。這道題wa了半天，居然是我忘記把文件輸入輸出那句話給刪了，尼瑪！下面是代碼：
#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'){//搜到終點 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]=='.'){//如果在迷宮內，並且未走過 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不為1，那就要在繼續往後走 } 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'){//找到起點 sx=i; sy=j; sz=k; } } memset(vis,0,sizeof(vis)); bfs(sx,sy,sz); } return 0; } #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'){//搜到終點
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]=='.'){//如果在迷宮內，並且未走過
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不為1，那就要在繼續往後走
}
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'){//找到起點
sx=i;
sy=j;
sz=k;
}
}
memset(vis,0,sizeof(vis));
bfs(sx,sy,sz);
}
return 0;
}
