POJ 2251 bfs

DESCRIPTION:
给你一个三维的迷宫。问你是否能从起点走到终点。如果能，输出最小步数。对我来说难得就是我没有想到怎么把他给你的三维图转换成map。恩。、好像解题报告上说。只要是这种的最短路都要用bfs。用dfs回很难。不太懂耶。>_<...

然后就是普通的bfs了。然后忘了三个输入全为0的时候结束程序。然后WA了一会。。然后就没有然后了。233333333333

附代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;

int l, r, c;
int ans;
bool used[40][40][40];
bool map[40][40][40];

struct Node
{
int l, r, c;
int step;
Node()
{
step = 0;
}
}node[30000], now, temp, st, ed;

int move[6][3] = {0, 0, 1, 0, 0, -1, 0, 1, 0, 0, -1, 0, 1, 0, 0, -1, 0, 0};

bool check(int l, int r, int c)
{
if (l >= 0 && l < 30 && r >= 0 && r < 30 && c >= 0 && c < 30 && !used[l][r][c] && map[l][r][c])
return true;
return false;
}

bool bfs(int i, int j, int k)
{
int top = 0;
int tail = 0;
node[tail++] = st;
used[st.l][st.r][st.c] = 1;
while(top < tail)
{
now = node[top++];
if (now.l == ed.l && now.r == ed.r && now.c == ed.c)
{
ans = now.step;
return true;
}
for (int i=0; i<6; ++i)
{
temp.l = now.l + move[i][0];
temp.r = now.r + move[i][1];
temp.c = now.c + move[i][2];
if (check(temp.l, temp.r, temp.c))
{
temp.step = now.step + 1;
node[tail++] = temp;
used[temp.l][temp.r][temp.c] = true;
}
}
}
return false;
}

int main()
{
char t;
while(cin >> l)
{
cin >> r >> c;
if (l == 0 && r == 0 && c == 0)
break;
memset(used, 0, sizeof(used));
memset(map, 0, sizeof(map));
for (int ll=0; ll<l; ++ll)
{
for (int rr=0; rr<r; ++rr)
{
for (int cc=0; cc<c; ++cc)
{
cin >> t;
if (t == 'S')
{
map[ll][rr][cc] = true;
st.l = ll;
st.r = rr;
st.c = cc;
st.step = 0;
}
if (t == '.')
map[ll][rr][cc] = true;
if (t == 'E')
{
map[ll][rr][cc] = true;
ed.l = ll;
ed.r = rr;
ed.c = cc;
}
}
}
}
if (bfs(st.l, st.r, st.c))
cout << "Escaped in " << ans << " minute(s)." << endl;
else cout << "Trapped!\n";
}
return 0;
}