hdu 1116 并查集判断欧拉回路通路

判断一些字符串能首尾相连连在一起

并查集求欧拉回路和通路

Sample Input

3
2
acm
ibm
3
acm
malform
mouse
2
ok
ok

Sample Output

The door cannot be opened.
Ordering is possible.
The door cannot be opened.

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m,tt;
 char s[MAXN];
 int in[],out[],vis[],f[],p[];
 int find(int x)
 {
     if(f[x]==-)return x;
     return f[x]=find(f[x]);
 }
 void bing(int x,int y)
 {
     int t1=find(x);
     int t2=find(y);
     if(t1!=t2)
     {
         f[t1]=t2;
     }
 }
 void init()
 {
     cl(vis);
     cl(in);
     cl(out);
     memset(f,-,sizeof(f));
 }
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     scanf("%d",&tt);
     while(tt--)
     {
         init();
         scanf("%d",&n);
         for(i=;i<n;i++)
         {
             scanf("%s",s);
             int len=strlen(s);
             int x=s[]-'a';
             int y=s[len-]-'a';
             out[x]++;
             in[y]++;
             bing(x,y);
             vis[x]=vis[y]=;
         }
         int cnt=;  //统计个数
         for(i=;i<;i++)
         {
             if(vis[i]&&find(i)==i)
             {
                 cnt++;
             }
         }
         if(cnt>)
         {
             printf("The door cannot be opened.\n");
             continue;
         }
         int tot=;  //统计出入度不相同的点
         for(i=;i<;i++)
         {
             if(vis[i]&&in[i]!=out[i])
             {
                 p[tot++]=i;
             }
         }
         if(tot==)
         {
             printf("Ordering is possible.\n");
             continue;
         }
         if(tot==&&((out[p[]]-in[p[]]==&&in[p[]]-out[p[]]==)||(out[p[]]-in[p[]]==&&in[p[]]-out[p[]]==))) //欧拉通路
         {
             printf("Ordering is possible.\n");
             continue;
         }
         printf("The door cannot be opened.\n");
     }
 }