反片语 Ananagrams，UVa156

Ananagrams

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

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

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.

Sample input

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

Sample output

DiskNotEderaildrIedeyeladdersoon

题意：输入一些单词,找出所有满足如下条件的单词：该单词不能通过字母重排，得到输入文本的另外一个单词。在判断是否满足条件时，字母不分大小写，但在输出时应保留输入中的大小写，按字典序进行排序。

思路：先输入单词，（利用vector），先把单词中的大写全部转化成小写，然后把单词中的每个字母都排序一遍，然后放进<map>里进行统计。如果只出现一次就输出。

知识点：

count方法：

统计map中某个键值出现的次数，因为map中键值唯一，所以此方法可以用来检测某键值是否存在，例如在删除时可以phone.count(name)，若为0则可以提示用户此键值不存在，若为1则直接删除。

1. #include <iostream>  
2. #include <cstring>  
3. #include <cctype>  
4. #include <vector>  
5. #include <map>  
6. #include <algorithm>  
7. using namespace std;  
8.   
9. map<string,int> cnt;  
10. vector<string> words;  
11.   
12. string repr(const string& s) //把单词标准化，即大写变小写，排序  
13. {  
14.     string ans=s;  //借用第三方变量进行转换，使最终输出时还能保留大写部分。  
15.     for(int i=0;i<ans.length();i++) ans[i]=tolower(ans[i]); //ps：toupper()是将小写转大写  
16.     sort(ans.begin(),ans.end());  
17.     return ans;  
18. }  
19. int main()  
20. {  
21.     int n=0;string s;  
22.     while(cin>>s)  
23.     {  
24.         if(s[0]=='#') break;  
25.         words.push_back(s); //存入vector  
26.         string r=repr(s);  
27.         if(!cnt.count(r)) cnt[r]=0;  
28.         cnt[r]++;  
29.     }  
30.     vector<string> ans;  
31.     for(int i=0;i<words.size();i++)  
32.         if(cnt[repr(words[i])]==1) ans.push_back(words[i]);  
33.     sort(ans.begin(),ans.end());  
34.     for(int i=0;i<ans.size();i++) cout<<ans[i]<<endl;  
35.     return 0;  
36. }