题意：有n个只包含两个字母的字符串， 要求构造一个m*m的字母矩阵， 使得矩阵的每行每列都不包含所给的字符串， m要尽量大，
如果大于20的话构造20*20的矩阵就行了。
析：开始吧，并没有读对题意，后来才看懂什么意思，然后主要思想就是如果有环，那么一定是可以构造成20*20的，只要环一直重复就好，如果没有环，
那么就要找最长的链，然后矩阵长宽就是 （max+1）/2，然后和上面一样构造就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
using namespace std :: tr1; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 26 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int G[maxn][maxn];
char s[5];
int ans[maxn], a[maxn];
bool vis[maxn];
int cnt; bool dfs(int u, int num){
bool ok = false;
for(int i = 0; i < 26; ++i){
if(G[u][i]){
a[num] = i;
if(vis[i]){
cnt = num;
return true;
}
else{
vis[i] = true;
if(dfs(i, num+1)) return true;
vis[i] = false;
}
ok = true;
}
}
if(cnt < num && ok){
cnt = num;
memcpy(ans, a, sizeof ans);
}
return false;
} int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
memset(G, -1, sizeof G);
for(int i = 0; i < n; ++i){
scanf("%s", s);
G[s[0]-'a'][s[1]-'a'] = 0;
} int anss = 0;cnt = 0;
for(int i = 0; i < 26; ++i){
memset(vis, false, sizeof vis);
vis[i] = true;
a[0] = i;
if(dfs(i, 1)){
bool ok = false;
int id = 0;
for(int j = 0; j < cnt; ++j)
if(ok) ans[id++] = a[j];
else if(a[j] == a[cnt]){ ans[id++] = a[j]; ok = true; }
anss = id;
break;
}
} if(anss){
for(int i = 0; i < 20; ++i){
for(int j = 0; j < 20; ++j)
printf("%c", ans[(j+i)%anss]+'a');
printf("\n");
}
}
else{
int x = (cnt+2)/2;
for(int i = 0; i < x; ++i){
for(int j = 0; j < x; ++j)
printf("%c", ans[i+j]+'a');
printf("\n");
}
} }
return 0;
}