UVA 10129 Play on Words

欧拉回路

以字母为结点，单词为边；注意两个相同的单词表示两条边。

并查集判断是否连通，出度，入度判断是否是欧拉回路

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;

 int map[];
 int g[][];
 int num[][];
 int v[],f[];

 int father (int i){
     f[i]=f[i]==i?f[i]:father (f[i]);
     return f[i];
 }

 int main (){
     int t,n;
     scanf ("%d",&t);
     while (t--){
         cin>>n;
         char s[];
         memset (v,,sizeof v);
         memset (g,,sizeof g);
         for (int i=;i<n;i++){
             scanf ("%s",s);
             int len=strlen (s);
             g[s[]-'a'][s[len-]-'a']++;
             v[s[]-'a']=v[s[len-]-'a']=;
         }
         int flag=;
         for (int i=;i<;i++)  f[i]=i;
         for (int i=;i<;i++)
             for (int j=;j<;j++)
                 if (g[i][j]||g[j][i])
                     f[father(i)]=father(j);
         for (int i=;i<;i++){
             for (int j=;j<;j++){//cout<<i<<j<<endl;
                 if (v[i]&&v[j]){
                     if (father(i)!=father(j)){
                             flag=;break ;
                     }
                 }
             }
             if (!flag)
                 break ;
         }
         if (!flag){
             printf ("The door cannot be opened.\n");
             continue ;
         }
         int c,r;//cout<<"error"<<endl;
         c=;r=;
         for (int i=;i<;i++){
             int temp=;
             for (int j=;j<;j++){
                 temp+=g[i][j]-g[j][i];
             }
             if (temp==&&c)
                 c=;
             else if (temp==-&&r)
                 r=;
             else if (temp==) ;
             else {
                 flag=;
                 break ;
             }
         }
         if (flag)
             printf ("Ordering is possible.\n");
         else printf ("The door cannot be opened.\n");
     }
     return ;
 }