UVa Problem 850 Crypt Kicker II （解密 II）

// Crypt Kicker II （解密 II）// PC/UVa IDs: 110304/850, Popularity: A, Success rate: average Level: 2// Verdict: Accepted// Submission Date: 2011-05-23// UVa Run Time: 0.016s//// 版权所有（C）2011，邱秋。metaphysis # yeah dot net#include <iostream>#include <vector>#include <sstream>#include <iterator>using namespace std;void decipher(vector < string > & plain, vector < vector < string > > & encrypted){string cipher(26, '*');for (vector < vector < string > >::iterator t = encrypted.begin(); t != encrypted.end(); t++)// 密文单词数量和明文单词数量相等才继续比较。if ((*t).size() == plain.size()){// 密文各单词长度与明文各单词长度相同才继续予以比较。vector < string >::iterator a = (*t).begin();vector < string >::iterator b = plain.begin();for (; a != (*t).end() && (*a).length() == (*b).length(); a++, b++);// 各单词长度相同，尝试匹配。if (a == (*t).end()){// 保存当前密码，以便不能匹配时恢复。string tmp = cipher;bool matched = true;for (a = (*t).begin(), b = plain.begin(); a != (*t).end(); a++, b++){string e = *a;string p = *b;for (int i = 0; i < e.length(); i++)// 若当前密文字母无对应明文文字母匹配，// 则将密文字母与明文字母匹配。if (cipher[e[i] - 'a'] == '*'){cipher[e[i] - 'a'] = p[i];}else{// 若已有明文字母与该密文字母// 相匹配，则检查需要// 匹配的密文是否与当前相同，// 若不同，则表示有冲突。if (cipher[e[i] - 'a'] != p[i]){matched = false;break;}}if (!matched)break;}// 满足所有匹配，按当前cipher输出密文，否则还原密码为初始状态。if (matched){for (vector < vector < string > >::iterator t = encrypted.begin(); t != encrypted.end(); t++){for (vector < string >::iterator v = (*t).begin(); v != (*t).end(); v++){string tmp = (*v);for (int j = 0; j < tmp.length(); j++)cout << cipher[tmp[j] - 'a'];if (v < (*t).end() - 1)cout << " ";}cout << endl;}return;}elsecipher = tmp;}}// 无解。cout << "No solution." << endl;}int main(int ac, char *av[]){vector < vector < string > > encrypted;// 保存密文。vector < string > plain;// 已知明文。string word, line;// 当前单词，当前行。int cases;// 测试数据组数。int current = 0;// 当前数据组数。// 将明文拆分成单词。istringstream iss("the quick brown fox jumps over the lazy dog");while (iss >> word)plain.push_back(word);// 读入数据组数。cin >> cases;cin.ignore();// 忽略空行。getline(cin, line);while (current < cases){encrypted.clear();// 读入密文，并拆分成单词。while (getline(cin, line), line.length() > 0){// 必须将输入字符串流的标志清空才能进行新的输入。iss.clear();iss.str(line);vector < string > tmp;while (iss >> word)tmp.push_back(word);encrypted.push_back(tmp);}// 解码。decipher(plain, encrypted);current++;if (current < cases)cout << '\n';}return 0;}