题意：给定 n 个 字符串，让你构造出一个最短，字典序最小的字符串，包括这 n 个字符串。
析：首先使用状压DP，是很容易看出来的，dp[s][i] 表示已经满足 s 集合的字符串以 第 i 个字符串结尾，他很容易就求得最短长度，但是这个字符串怎么构造呢，
由于要字典序最小，所以就不好搞了，挺麻烦的，所以我们利用贪心的思路，我们可以这样定义，dp[s][i] 表示已经满足 s 集合的字符串以 第 i 个字符串开头，
从后向前放，状态转移方程为：dp[s|(1<<i)][i] = min{ dp[s][j] + dist[k][j] }，dist[k][j] 表示把 k 放到 j 前面所要的最短长度，这个数组我们可以通过预处理来得到，
注意这里是把 k 放到 j 的前面，不是把 j 放到 k 的后面。最后还要注意把包含的字符串去掉，还有如果全为一样的情况，要注意特殊判断，这个点我RE了一晚上。。。
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 <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 100000000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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 bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[1<<16][16], dist[20][20];
string str[20], ans;
vector<string> v;
void init(){
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j) if(i != j){
int t = min(v[i].size(), v[j].size());
dist[i][j] = 0;
for(int k = t; k >= 0; --k)
if(v[i].substr(v[i].size() - k) == v[j].substr(0, k)){
dist[i][j] = v[i].size() - k;
break;
}
}
}
void dfs(int id, int s){
if(s == 0) return ;
string ss = "Z";
int x;
for(int i = 0; i < n; ++i){
if(!(s&(1<<i))) continue;
if(dp[s|(1<<id)][id] == dp[s][i] + dist[id][i]){
int xx = v[id].size() - dist[id][i];
string sss = v[i].substr(xx);
if(ss > sss) ss = sss, x = i;
}
}
ans += ss;
dfs(x, s^(1<<x));
}
int main(){
ios_base::sync_with_stdio(false);
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
cin >> n;
cout << "Scenario #" << kase << ":" << endl;
for(int i = 0; i < n; ++i) cin >> str[i];
v.clear();
for(int i = 0; i < n; ++i){
bool ok = true;
for(int j = 0; j < n; ++j){
if(i == j || str[i].size() > str[j].size()) continue;
if(str[j].find(str[i]) != string::npos){
ok = false;
break;
}
}
if(ok) v.push_back(str[i]);
}
if(v.empty()){
cout << str[0] << endl << endl;
continue;
}
sort(v.begin(), v.end());
n = v.size();
init();
int all = 1<<n;
memset(dp, INF, sizeof dp);
for(int i = 0; i < n; ++i)
dp[1<<i][i] = v[i].size();
for(int i = 0; i < all; ++i)
for(int j = 0; j < n; ++j){
if(dp[i][j] == INF) continue;
for(int k = 0; k < n; ++k){
if(i & (1<<k)) continue;
dp[i|(1<<k)][k] = min(dp[i|(1<<k)][k], dp[i][j] + dist[k][j]);
}
}
int id = 0;
for(int i = 0; i < n; ++i) if(dp[all-1][id] > dp[all-1][i]) id = i;
ans = v[id];
dfs(id, (all-1)^(1<<id));
cout << ans << endl << endl;
}
return 0;
}