hdu1116

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<string.h>
 4 #include<stdlib.h>
 5 #include<iostream>
 6 using namespace std;
 7 const int N=30;
 8 int father[N],vis[N];
 9 int find(int x)//查找根
10 {
11     if(father[x]!=x)
12     father[x]=find(father[x]);
13     return father[x];
14 }
15 void make(int a,int b)//合并操作
16 {
17     int f1=find(a);
18     int f2=find(b);
19     if(f1!=f2)
20     father[f1]=f2;
21 }
22 
23 int main()
24 {
25     //freopen("in.txt","r",stdin);
26     int t;
27     cin>>t;
28     while(t--)
29     {
30         int out[N],in[N],p[N];
31         memset(in,0,sizeof(in));
32         memset(out,0,sizeof(out));
33         memset(vis,0,sizeof(vis));
34         int n;
35         for(int i=0;i<26;i++)
36         father[i]=i;//数组赋初值
37         cin >> n;
38         while(n--)
39         {
40             char str[1005];
41             cin>>str;
42             int a=str[0]-‘a‘;
43             int b=str[strlen(str)-1]-‘a‘;
44             out[a]++;
45             in[b]++;
46             make(a,b);//合并操作
47             vis[a]=vis[b]=1;
48         }
49         for(int i=0;i<26;i++)
50         father[i]=find(i);//寻找每个点的根节点
51         int cnt=0;
52         for(int i=0;i<26;i++)
53         {
54             if(vis[i] && father[i]==i)//确定有几颗树
55             cnt++;
56         }
57         if(cnt>1)
58         {
59             printf("The door cannot be opened.\n");
60             continue;
61         }
62         int j=0;
63         for(int i=0;i<26;i++)
64         {
65             if(vis[i] && out[i]!=in[i])//找起点和终点
66             p[j++]=i;
67         }
68         if(j==0)//环
69         {
70             printf("Ordering is possible.\n");
71             continue;
72         }
73         if(j==2 && ((out[p[0]]-in[p[0]]==1 && in[p[1]]-out[p[1]]==1 ) ||
74                     (in[p[0]]-out[p[0]]==1 && out[p[1]]-in[p[1]]==1 ) ) )
75         {//起点和终点的判断
76             printf("Ordering is possible.\n");
77             continue;
78         }
79         printf("The door cannot be opened.\n");
80     }
81     return 0;
82 }