题意：给定一篇文章，将每个单词的首尾字母不变，中间顺序打乱，然后将单词之间的空格去掉，得到一个序列，给出一个这样的序列，给你一个字典，将原文翻译出来。
析：在比赛的时候读错题了，忘记首尾字母不变了，一直WA。暴力求解，去深搜每个单词，做一些恰当的优化，能不进行的就不进行。胡搞的。
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 <unordered_map>
//#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 = 1e4 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};
const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 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;
}
struct Node{
char s[105];
int id, n;
};
struct node{
string s;
char pre, last;
node(string ss, char p, char l) : s(ss), pre(p), last(l) { }
};
unordered_map<string, vector<Node> > mp;
bool vis[105];
vector<int> v;
vector<node> ans;
int cnt;
string str;
int solve(const string &s, const string &ss){
int ans = 0;
for(int i = 0; i < mp[s].size(); ++i)
if(mp[s][i].s[0] == ss[0] && mp[s][i].s[mp[s][i].n-1] == ss[ss.size()-1]) ++ans;
return ans;
}
bool dfs(int cur){
if(cur >= str.size()){ ++cnt; return true; }
if(cnt > 1) return true;
bool ok = false;
for(int i = 0; i < v.size() && v[i]+cur <= str.size(); ++i){
string s = str.substr(cur, v[i]);
if(s.size() > 2){
string ss = s.substr(1, s.size()-2);
sort(ss.begin(), ss.end());
if(!mp.count(ss)) continue;
int t = solve(ss, s);
if(!t) continue;
if(dfs(cur+v[i])){
if(t == 1){ ok = true; ans.push_back(node(ss, s[0], s[s.size()-1])); }
else { cnt = 5; return true; }
}
}
else{
if(!mp.count(s)) continue;
int t = 0;
for(int j = 0; j < mp[s].size(); ++j)
if(mp[s][j].n == s.size()) ++t;
if(!t) continue;
if(dfs(v[i]+cur)){
if(t == 1){ ok = true; ans.push_back(node(s, 0, s.size())); }
else { cnt = 5; return true; }
}
}
}
return ok;
}
int main(){
int T; cin >> T;
while(T--){
cin >> str;
scanf("%d", &m);
mp.clear();
ans.clear();
v.clear();
memset(vis, false, sizeof vis);
char t[105];
Node u;
for(int i = 0; i < m; ++i){
scanf("%s", u.s);
int n = strlen(u.s);
u.n = n;
if(n > 2){
memcpy(t, u.s+1, n-2);
t[n-2] = 0;
sort(t, t+n-2);
u.id = mp[t].size();
mp[t].push_back(u);
}
else{
memcpy(t, u.s, n+1);
u.id = 0;
mp[t].push_back(u);
}
vis[n] = true;
}
for(int i = 1; i < 105; ++i) if(vis[i]) v.push_back(i);
cnt = 0; dfs(0);
if(cnt > 1) puts("ambiguous");
else if(!cnt) puts("impossible");
else {
for(int i = ans.size()-1; i >= 0; --i){
if(i != ans.size()-1) putchar(' ');
node &anss = ans[i];
for(int j = 0; j < mp[anss.s].size(); ++j){
if(anss.pre == 0 && mp[anss.s][j].n == anss.last){ printf("%s", mp[anss.s][j].s); break; }
else if(anss.pre == mp[anss.s][j].s[0] && anss.last == mp[anss.s][j].s[mp[anss.s][j].n-1]){
printf("%s", mp[anss.s][j].s); break;
}
}
}
printf("
");
}
}
return 0;
}