Codeforces Gym 100286J Javanese Cryptoanalysis 傻逼暴力

原题地址：http://codeforces.com/gym/100286/attachments/download/2013/20082009-acmicpc-northeastern-european-regional-contest-neerc-08-en.pdf

此题题意是给你一个单对单密文，让你还原为原文，原文有个性质是，每个单词都是元音和辅音交替组成。

做法是直接5重for，暴力枚举AEIOU分别对应的字母，然后检查，然后输出

详见代码：

//#include<iostream>
#include<fstream>
#include<vector>
#include<string>
#include <string.h>

#define MAX_A 26
#define MAX_N 100005
using namespace std;

vector<int> G[MAX_A];

string str[MAX_N];
int n = ;
string s;

bool g[MAX_A][MAX_A];

int vo[MAX_A];
int tot = ;
bool flag = ;

int main() {
    ifstream cin("javanese.in");
    ofstream cout("javanese.out");
    cin.sync_with_stdio(false);
    for (int i = ; i < ; i++) {
        if (i !=  && i != 'E' - 'A' && i != 'I' - 'A' && i != 'O' - 'A' && i != 'U' - 'A')
            vo[tot++] = i;
    }
    while (cin >> s) {
        if (s == "*")break;
        str[n++] = s;
        for (int i = ; i +  < s.length(); i++) {
            int u = s[i] - 'A', v = s[i + ] - 'A';
            if (u == v)flag = false;
            if (g[u][v])continue;
            G[s[i] - 'A'].push_back(s[i + ] - 'A');
            g[u][v] = ;
        }
        if (flag == ) {
            cout << "impossible" << endl;
            return ;
        }
    }
    int a[];

    for (a[] = ; a[] < ; a[]++)
        for (a[] = a[] + ; a[] < ; a[]++)
            for (a[] = a[] + ; a[] < ; a[]++)
                for (a[] = a[] + ; a[] < ; a[]++)
                    for (a[] = a[] + ; a[] < ; a[]++) {
                        int f[];
                        memset(f, -, sizeof(f));
                        int tmp = ;
                        bool t = ;
                        for (int i = ; i <  && t; i++)
                            for (int j = ; j <  && t; j++)if (g[a[i]][a[j]])t = ;
                        if (t == )continue;
                        for (int i = ; i < ; i++)
                            for (int j = ; j < ; j++)
                                if (g[i][j] && i != a[] && i != a[] && i != a[] && i != a[] && i != a[] &&
                                    j != a[] && j != a[] && j != a[] && j != a[] && j != a[])
                                    t = ;
                        if (t == )continue;
                        for (int i = ; i < n; i++, cout << " ")
                            for (int j = ; j < str[i].length(); j++) {
                                int u = str[i][j] - 'A';
                                if (f[u] != -) {
                                    cout << (char)(f[u] + 'A');
                                    continue;
                                }
                                if (u == a[])f[u] = 'A' - 'A';
                                else if (u == a[])f[u] = 'E' - 'A';
                                else if (u == a[])f[u] = 'I' - 'A';
                                else if (u == a[])f[u] = 'O' - 'A';
                                else if (u == a[])f[u] = 'U' - 'A';
                                else f[u] = vo[tmp++];
                                cout << (char)(f[u] + 'A');
                            }
                        cout << endl;
                        return ;
                    }
    cout << "impossible" << endl;
    return ;
}