这题要是有BT数据，可以有10^10个点，输入都读不完就铁定TLE
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>

using namespace std;

int cnt;
int n;
map<long long int, int> idx;
int adjNum[10000];
int adj[10000][10];
int qd, zd;

int getIdx(){
	int Temp;
	scanf("%d",&Temp);
	long long int temp = Temp;
	if(temp == -1) return -1;
	for(int i = 1; i < n; i++){
		int tmp;
		scanf("%d", &tmp);
		temp *= 10;
		temp += tmp;
	}
	map<long long int, int>::iterator it = idx.find(temp);
	if(it != idx.end()) return it->second;
	idx.insert(pair<long long int, int>(temp, cnt));
	cnt ++;
	return cnt-1;
}

bool dfs(int q, bool *used){
	used[q] = 1;
	if(q == zd) return 1;
	for(int i = 0; i < adjNum[q]; i++){
		if(used[adj[q][i]]) continue;
		if(dfs(adj[q][i], used)) return 1;
	}
	return 0;
}

int main(int argc, char **argv){
	int ct = 0;
	while(1){
		cin >> n;
		if(!n) break;
		ct++;
		cout << "Maze #" << ct << " can";
		cnt = 0;
		idx.clear();
		for(int i = 0; i < 10000; i++) adjNum[i] = 0;
		qd = getIdx(), zd = getIdx();
		int x,y;
		while((x = getIdx()) != -1){
			y = getIdx();
			adj[x][adjNum[x]] = y;
			adjNum[x]++;
			adj[y][adjNum[y]] = x;
			adjNum[y]++;
		}
		bool used[10000] = {0};
		if(!dfs(qd, used)) cout << "not";
		cout << " be travelled" << endl;
	}
		
	return 0;
}