pku 1386 Play on Words 欧拉回路

pku 1386 Play on Words 欧拉回路

#include  < iostream >
using   namespace  std;
const   int  maxn = 1005 ;
const   int  M = 28 ;
int  map[M][M];
int  chu[M],ru[M];
bool  vis[M];        
int  n;
int  sum;

void  dfs( int  s)
{
    vis[s]=true;
    sum++;
    int i;
    for(i=0;i<26;i++)
        if(map[s][i]&&!vis[i])
            dfs(i);
}

int  main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(map,0,sizeof(map));
        memset(chu,0,sizeof(chu));
        memset(ru,0,sizeof(ru));
        memset(vis,0,sizeof(vis));

        scanf("%d",&n);
        int i;
        for(i=0;i<n;i++)
        {
            char word[maxn];
            scanf("%s",&word);
            int len=strlen(word);
            map[word[0]-'a'][word[len-1]-'a']=1;
            chu[word[0]-'a']++;
            ru[word[len-1]-'a']++;
        }
        int n0=0,n1=0,n2=0,n3=0;
        int s=-1,e=-1;
        for(i=0;i<26;i++)
        {
            
            if(chu[i]==0&&ru[i]==0)
                continue;
            n0++;
            if((ru[i]-chu[i])==1)
                n2++;
            if((chu[i]-ru[i])==0)
                n3++;
            if((chu[i]-ru[i])==1)
            {
                s=i;
                n1++;    
            }
            e=i;
        }
        sum=0;
        if(s!=-1)
            dfs(s);
        else
            dfs(e);
        if(sum!=n0)
        {
            printf("The door cannot be opened.\n");
            continue;
        }
        if( (n1==1&&n2==1&&n3==(n0-2)) || (n0==n3) )
            printf("Ordering is possible.\n");
        else
            printf("The door cannot be opened.\n");
    }
    return 0;
}