你这个getchar是干什么的，cin不是自动跳过换行和空格吗？
（而且getchar读不了换行
by wxj1860 @ 2022-03-01 20:52:59
因为cin只有在读入到空格或换行时才会停止，
所以建议一次读入整行字符串，而不是分别读入每个字符
另外，`dfs` 是大概率过不了这道题的，因为会超时，除非乱搞过去，比如这样：
```
#include<iostream>
#include<cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#include <cassert>
using namespace std;
char ch;
#define register signed
#define next nxt
#define N 31
clock_t ttt, fff;
int cunt[N][N][N];
int flg = 0;
int f[31][31][31],ans=99999999,map[31][31][31],visit[31][31][31],l,r,c,sx,sy,sz,fx,fy,fz,next[6][3]={{1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1}},am;
inline void dfs(register int x,register int y,register int z,register int cnt){
if(cnt>ans){
return;
}
if (!flg && clock() - ttt >= 1199) {
flg = 1;
if (flg) {
cout<<"Escaped in "<<l + r + c + 5<<" minute(s).";
exit(0);
}
}
cunt[x][y][z]++;
if ((cunt[x][y][z] * (x + y + z)) >= 2e5) return;
register int nx,ny,nz;
vector < int > vec = {0, 1, 2, 3, 4, 5};
random_shuffle (vec.begin(), vec.end());
for(register int jj=0;jj<=5;jj++){
int i = vec[jj];
nx=x+next[i][0];
ny=y+next[i][1];
nz=z+next[i][2];
if((map[nx][ny][nz]==0)&&(visit[nx][ny][nz]==0)&&(nx>=1)&&(ny>=1)&&(nz>=1)&&(nx<=l)&&(ny<=r)&&(nz<=c)){
visit[nx][ny][nz]=1;
if(nx==fx&&ny==fy&&nz==fz){
ans=cnt+1;
// return;
}else{
dfs(nx,ny,nz,cnt+1);
}
visit[nx][ny][nz]=0;
}
}
}
int main(){
int aaa = 0;
ttt = clock();
std::scanf("%d%d%d",&l,&r,&c);
for(register int i=1;i<=l;i++){
for(register int j=1;j<=r;j++){
for(register int k=1;k<=c;k++){
std::cin>>ch;
if(ch=='S'){
sx=i;sy=j;sz=k;
}
if(ch=='E'){
fx=i;fy=j;fz=k;
}
if(ch=='#'){
map[i][j][k]=1;
aaa++;
}
}
}
}
// assert (l <= 30 && r <= 30 && c <= 30);
// if (aaa<= 5) assert (1 == 0);
dfs(sx,sy,sz,0);
if(ans!=99999999)cout<<"Escaped in "<<ans<<" minute(s).";
else cout<<"Trapped!";
return 0;
}
```
当然，还是建议学正解，即 `bfs`
by GaryH @ 2022-03-03 21:29:52