10129 play on words -- and query the set + Euler's path !!!

Judging whether all strings can be connected together is actually to judging whether the constructed directed graph has Euler's loop or Euclidean;

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include <algorithm>#define N 100010using namespace std;char c[N][2];int al[123];int in[27],out[27];int f[27],ccount;bool yo[26];int tou(int x){    if(f[x]!=x)    {        f[x]=tou(f[x]);    }    return f[x];}void build(int x,int y){    int toux=tou(x),touy=tou(y);    if(touy!=toux)    {        ++ccount;        f[toux]=touy;    }}int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        memset(al,0,sizeof(al));        memset(in,0,sizeof(in));        memset(out,0,sizeof(out));        memset(yo,0,sizeof(yo));        memset(f,0,sizeof(f));        scanf("%d%*c",&n);        int geshu=0;        for(int i=0; i<n; i++)        {            char cc,ccc;            scanf("%c",&c[i][0]);            if(!yo[c[i][0]-'a'])            {                yo[c[i][0]-'a']=1;                ++geshu;            }            while(scanf("%c",&cc)&&cc!='\n')            {                ccc=cc;            }            c[i][1]=ccc;            if(!yo[c[i][1]-'a'])            {                yo[c[i][1]-'a']=1;                ++geshu;            }            in[c[i][1]-'a']++;            out[c[i][0]-'a']++;        }        for(int i=0; i<26; i++)            f[i]=i;        ccount=0;        for(int i=0; i<n; i++)            build(c[i][0]-'a',c[i][1]-'a');        int count=0,flag=1;        if(ccount+1!=geshu)            flag=0;        if(flag)            for(int i=0; i<26; i++)            {                if(in[i]!=out[i])                {                    count++;                    if(abs(in[i]-out[i])!=1||count>2)                    {                        flag=0;                        break;                    }                }            }        if(!flag)            printf("The door cannot be opened.\n");        else            printf("Ordering is possible.\n");    }    return 0;}