题目大意：给出每一层的路况，'O'表示这路是可以走的，'X'表示此路不通，程序结果要求输出四方体的维数（已经给出）和起点到达终点所需的最小步数。（类似的题：hdu 1254 胜利大逃亡）
第二组数据
START 3 XXX //最下面一层 XXX XXX OOO //中间一层 OOO OOO XXX //最上面一层 XXX XXX 0 0 1 2 2 1 END
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
#define MAX 15
struct Node
{
int x, y, z;
int num;
Node(int x, int y, int z, int num):x(x),y(y),z(z), num(num){}
Node(){}
};
char n[MAX][MAX][MAX];
Node s, e;
int v[MAX][MAX][MAX], flag;
int dx[]={0,0,1,-1,0,0};
int dy[]={0,0,0,0,1,-1};
int dz[]={1,-1,0,0,0,0};
int main()
{
int N, x, y, z, i;
char str[10], c;
while (cin>>str>>N)
{
flag = 0;
memset(v, 0, sizeof(v));
for (z=1; z<=N; z++)
for (x=1; x<=N; x++)
for (y=1; y<=N; y++)
cin>>n[x][y][z];
cin>>s.x>>s.y>>s.z;
cin>>e.x>>e.y>>e.z;
cin>>str;
queue<Node> q;
v[s.x+1][s.y+1][s.z+1] = 1;
q.push(Node(s.x+1,s.y+1,s.z+1,0));
while (!q.empty())
{
Node tmp = q.front();
q.pop();
if (tmp.x==e.x+1&&tmp.y==e.y+1&&tmp.z==e.z+1)
{
flag = 1;
cout<<N<<" "<<tmp.num<<endl;
break;
}
//v[tmp.x][tmp.y][tmp.z] = 1;
for (i=0; i<6; i++)
{
if (v[tmp.x+dx[i]][tmp.y+dy[i]][tmp.z+dz[i]] == 0 && n[tmp.x+dx[i]][tmp.y+dy[i]][tmp.z+dz[i]]=='O')
{
q.push(Node(tmp.x+dx[i],tmp.y+dy[i],tmp.z+dz[i], tmp.num+1));
v[tmp.x+dx[i]][tmp.y+dy[i]][tmp.z+dz[i]]=1;
}
}
}
if (flag==0)
cout<<"NO ROUTE"<<endl;
}
return 0;
}