题意

单词链接, 如acm, malform, mouse可以链接

思路

欧拉回路

1.图连通

2.图中所有点的出度==入度，或只有两个奇度点并且一个点的入度比出度大1（终点），另一个点的出度比入度大1（起点）

调试的时候遇到几个比较坑的数据 需要特殊处理判断一下

input

3

2

aa

aaa

1

ab

2

aa

bb

output

Ordering is possible.

Ordering is possible.

The door cannot be opened.

AC代码

#include 
    
     #include 
     
      #include 
      
       using namespace std;int in[30], out[30], G[30][30], vis[30];void dfs(int a){    vis[a] = 1;    for (int b = 0; b < 26; b++)        if (!vis[b] && G[a][b])            dfs(b);}int main(){    int T, n;    char s[1000+10];    scanf("%d",&T);    while(T--){        bool ok = true, ok2 = false;        memset(vis, 0, sizeof(vis));        memset(G, 0, sizeof(G));        memset(in, 0, sizeof(in));        memset(out, 0, sizeof(out));        scanf("%d",&n);        if(n==1)    ok2 = true;        while(n--){            scanf("%s",s);            int len = strlen(s);            int a = s[0]-'a', b = s[len-1]-'a';            G[a][b]++;            in[b]++, out[a]++;        }        if(ok2){            puts("Ordering is possible.");            continue;        }        int cnt1 = 0, cnt2 = 0, cnt = 0;        for( int i = 0; i < 26; i++ ){            if( in[i] == out[i] )   continue;            if( in[i] + 1 == out[i] )   cnt1++;            else if( out[i] + 1 == in[i] )  cnt2++;            else{ ok = false;  break;}        }        if(cnt1 && cnt2 && cnt1+cnt2 > 2)   ok = false;        int num = 0;        if(ok){            for( int i = 0; i < 26; i++ )                if( out[i] ){                    dfs(i);                    break;                }            for( int i = 0; i < 26; i++ )                if( in[i] + out[i] )                    if( !vis[i] ){                        ok = false;                        break;                    }        }        if(ok) puts("Ordering is possible.");        else puts("The door cannot be opened.");    }    return 0;}