HDU 2340 Obfuscation (暴力)

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;
}