hdu1116

http://acm.hdu.edu.cn/showproblem.php?pid=1116

 #include<stdio.h>
 #include<math.h>
 #include<string.h>
 #include<stdlib.h>
 #include<iostream>
 using namespace std;
 const int N=;
 int father[N],vis[N];
 int find(int x)//查找根
 {
     if(father[x]!=x)
     father[x]=find(father[x]);
     return father[x];
 }
 void make(int a,int b)//合并操作
 {
     int f1=find(a);
     int f2=find(b);
     if(f1!=f2)
     father[f1]=f2;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     int t;
     cin>>t;
     while(t--)
     {
         int out[N],in[N],p[N];
         memset(in,,sizeof(in));
         memset(out,,sizeof(out));
         memset(vis,,sizeof(vis));
         int n;
         for(int i=;i<;i++)
         father[i]=i;//数组赋初值
         cin >> n;
         while(n--)
         {
             char str[];
             cin>>str;
             int a=str[]-'a';
             int b=str[strlen(str)-]-'a';
             out[a]++;
             in[b]++;
             make(a,b);//合并操作
             vis[a]=vis[b]=;
         }
         for(int i=;i<;i++)
         father[i]=find(i);//寻找每个点的根节点
         int cnt=;
         for(int i=;i<;i++)
         {
             if(vis[i] && father[i]==i)//确定有几颗树
             cnt++;
         }
         if(cnt>)
         {
             printf("The door cannot be opened.\n");
             continue;
         }
         int j=;
         for(int i=;i<;i++)
         {
             if(vis[i] && out[i]!=in[i])//找起点和终点
             p[j++]=i;
         }
         if(j==)//环
         {
             printf("Ordering is possible.\n");
             continue;
         }
         if(j== && ((out[p[]]-in[p[]]== && in[p[]]-out[p[]]== ) ||
                     (in[p[]]-out[p[]]== && out[p[]]-in[p[]]== ) ) )
         {//起点和终点的判断
             printf("Ordering is possible.\n");
             continue;
         }
         printf("The door cannot be opened.\n");
     }
     return ;
 }