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
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 int map[100010]; 7 int g[30][30]; 8 int num[26][2]; 9 int v[30],f[30]; 10 11 int father (int i){ 12 f[i]=f[i]==i?f[i]:father (f[i]); 13 return f[i]; 14 } 15 16 int main (){ 17 int t,n; 18 scanf ("%d",&t); 19 while (t--){ 20 cin>>n; 21 char s[2000]; 22 memset (v,0,sizeof v); 23 memset (g,0,sizeof g); 24 for (int i=0;i<n;i++){ 25 scanf ("%s",s); 26 int len=strlen (s); 27 g[s[0]-'a'][s[len-1]-'a']++; 28 v[s[0]-'a']=v[s[len-1]-'a']=1; 29 } 30 int flag=1; 31 for (int i=0;i<30;i++) f[i]=i; 32 for (int i=0;i<26;i++) 33 for (int j=0;j<26;j++) 34 if (g[i][j]||g[j][i]) 35 f[father(i)]=father(j); 36 for (int i=0;i<26;i++){ 37 for (int j=0;j<26;j++){//cout<<i<<j<<endl; 38 if (v[i]&&v[j]){ 39 if (father(i)!=father(j)){ 40 flag=0;break ; 41 } 42 } 43 } 44 if (!flag) 45 break ; 46 } 47 if (!flag){ 48 printf ("The door cannot be opened.\n"); 49 continue ; 50 } 51 int c,r;//cout<<"error"<<endl; 52 c=1;r=1; 53 for (int i=0;i<26;i++){ 54 int temp=0; 55 for (int j=0;j<26;j++){ 56 temp+=g[i][j]-g[j][i]; 57 } 58 if (temp==1&&c) 59 c=0; 60 else if (temp==-1&&r) 61 r=0; 62 else if (temp==0) ; 63 else { 64 flag=0; 65 break ; 66 } 67 } 68 if (flag) 69 printf ("Ordering is possible.\n"); 70 else printf ("The door cannot be opened.\n"); 71 } 72 return 0; 73 }
