学步园

大意略。

思路：有向图的欧拉路径。

条件：

1、连通

2、有2个奇点，其中出度比入度一个大1，起点，另一个小1，终点。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 50;

int ind[maxn], outd[maxn];

int vis[maxn];

int p[maxn];
int G[maxn][maxn];

char str[1010];

int n;

void init()
{
	memset(G, 0, sizeof(G));
	memset(vis, 0, sizeof(vis));
	memset(ind, 0, sizeof(ind));
	memset(outd, 0, sizeof(outd));
	for(int i = 0; i < 27; i++) p[i] = i;
}

int find(int x)
{
	return x == p[x]? x: p[x] = find(p[x]);
}

void Union(int a, int b)
{
	int x = find(a), y = find(b);
	if(x == y) return ;
	p[x] = y;
}

int check()
{
	int s = 0;
	for(int i = 0; i < 27; i++) if(vis[i] && p[i] == i) s++;
	return s == 1;
}

void read_case()
{
	init();
	scanf("%d", &n);
	while(n--)
	{
		scanf("%s", str);
		int u = str[0]-'a', v = str[strlen(str)-1]-'a';
		G[u][v] = 1;
		ind[v]++, outd[u]++;
		Union(u, v);
		vis[u] = vis[v] = 1;
	}
}

void solve()
{
	int flag = 1;
	read_case();
	if(!check()) flag = 0;
	int c1 = 0, c2 = 0;
	for(int i = 0; i < 27; i++) if(vis[i])
	{
		if(outd[i] != ind[i])
		{
			if(outd[i] - ind[i] == 1) c1++;
			else if(ind[i] - outd[i] == 1) c2++;
			else { flag = 0; break; }
		}
	}
	if(c1+c2 > 2) flag = 0;
	if(flag) printf("Ordering is possible.\n");
	else printf("The door cannot be opened.\n");
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		solve();
	}
	return 0;
}