Most crossword puzzle fans are used to anagrams--groups of words with the same letters in different orders--for example OPTS, SPOT, STOP, POTS and POST. Some words however do not have this attribute, no matter how you rearrange their letters, you cannot form another word. Such words are called ananagrams, an example is QUIZ.

 

Obviously such definitions depend on the domain within which we are working; you might think that ATHENE is an ananagram, whereas any chemist would quickly produce ETHANE. One possible domain would be the entire English language, but this could lead to some problems. One could restrict the domain to, say, Music, in which case SCALE becomes a relative ananagram (LACES is not in the same domain) but NOTE is not since it can produce TONE.

 

Write a program that will read in the dictionary of a restricted domain and determine the relative ananagrams. Note that single letter words are, ipso facto, relative ananagrams since they cannot be ``rearranged‘‘ at all. The dictionary will contain no more than 1000 words.

Input will consist of a series of lines. No line will be more than 80 characters long, but may contain any number of words. Words consist of up to 20 upper and/or lower case letters, and will not be broken across lines. Spaces may appear freely around words, and at least one space separates multiple words on the same line. Note that words that contain the same letters but of differing case are considered to be anagrams of each other, thus tIeD and EdiT are anagrams. The file will be terminated by a line consisting of a single #.

 

Output will consist of a series of lines. Each line will consist of a single word that is a relative ananagram in the input dictionary. Words must be output in lexicographic (case-sensitive) order. There will always be at least one relative ananagram.

ladder came tape soon leader acme RIDE lone Dreis peat ScAlE orb eye Rides dealer NotE derail LaCeS drIed noel dire Disk mace Rob dries #

Disk 

NotE 

derail 

drIed 

eye 

ladder 

soon

 1 # include <iostream>//输入输出流
 2 # include <string>
 3 # include <cctype> //包含 tolower 函数
 4 # include <vector>  //不定长数组
 5 # include <map>     //映射，在本例中36行有体现
 6 # include <algorithm> //一些函数，sort等
 7 using namespace std;
 8 
 9 map<string, int> cnt; //定义映射
10 vector<string> words; //一个叫words的不定长数组
11 
12 //将单词s进行标准化
13 string repr(const string& s)//声明变量s
14 {
15     string ans = s;//定义字符串类型数组，并使用第三方变量进行转换，并最终输出时还能保留大写部分
16     for (int i = 0; i < ans.length(); i++)
17         ans[i] = tolower(ans[i]);//全部转化成小写字母
18     sort(ans.begin(), ans.end());//从小到大按字典序进行排序
19     return ans;
20 }
21 
22 int main()
23 {
24     int n = 0;
25     string s;
26     while (cin >> s)
27     {
28         if (s[0] == ‘#‘) break;//遇到‘#’结束循环
29         words.push_back(s);//存入vector
30         string r = repr(s);//给所有单词进行排列，上面13行用“引用”的原因，方便调用
31         if (!cnt.count(r)) cnt[r] = 0; //!cnt.count(r) 的值，不是0就是1
32         cnt[r]++;      //count 是返回容器中r出现的次数，统计重新组成新单词的个数
33     }
34     vector<string> ans;//又重新复制了一份，用另外一个存储
35     for (int i = 0; i < words.size(); i++)
36       if (cnt[repr(words[i])] == 1)//此时用到了“映射”，如果不是映射的话，一个是字符串一个是int型，不能用等号的；找到唯一的单词
37             ans.push_back(words[i]);
38     sort(ans.begin(), ans.end());//排序
39     for (int i = 0; i < ans.size(); i++)
40         cout << ans[i] << "\n";
41 
42     getchar();
43     getchar();
44     
45     return 0;
46 }