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[100010];
int g[30][30];
int num[26][2];
int v[30],f[30];

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[2000];
        memset (v,0,sizeof v);
        memset (g,0,sizeof g);
        for (int i=0;i<n;i++){
            scanf ("%s",s);
            int len=strlen (s);
            g[s[0]-'a'][s[len-1]-'a']++;
            v[s[0]-'a']=v[s[len-1]-'a']=1;
        }
        int flag=1;
        for (int i=0;i<30;i++)  f[i]=i;
        for (int i=0;i<26;i++)
            for (int j=0;j<26;j++)
                if (g[i][j]||g[j][i])
                    f[father(i)]=father(j);
        for (int i=0;i<26;i++){
            for (int j=0;j<26;j++){//cout<<i<<j<<endl;
                if (v[i]&&v[j]){
                    if (father(i)!=father(j)){
                            flag=0;break ;
                    }
                }
            }
            if (!flag)
                break ;
        }
        if (!flag){
            printf ("The door cannot be opened.
");
            continue ;
        }
        int c,r;//cout<<"error"<<endl;
        c=1;r=1;
        for (int i=0;i<26;i++){
            int temp=0;
            for (int j=0;j<26;j++){
                temp+=g[i][j]-g[j][i];
            }
            if (temp==1&&c)
                c=0;
            else if (temp==-1&&r)
                r=0;
            else if (temp==0) ;
            else {
                flag=0;
                break ;
            }
        }
        if (flag)
            printf ("Ordering is possible.
");
        else printf ("The door cannot be opened.
");
    }
    return 0;
}