Uva-156 Ananagrams

Descripition  

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 drIed
noel dire Disk mace Rob dries
#
Sample Output
Disk
NotE
derail
drIed
eye
ladder

soon

这个题的大意是把所有不能满足重排后得到另外一个文本内单词的单词进行字典序排序，且区分大小写，实现重排即所有字母个数相同，实现重排辨别重复计数可以用映射内的count函数。

ac代码：

#include <bits/stdc++.h>using namespace std;string re(const string &s){    string ans = s;    for(int i = 0; i < ans.length(); i++)        ans[i] = tolower(ans[i]);    sort(ans.begin(),ans.end());    return ans;                           //使每个单词内的字母字典序排序}int main(){    map<string,int> cnt;    vector<string> words;    string s;    int n = 0;    while(cin>>s)    {        if(s[0] == '#')            break;        words.push_back(s);        string r = re(s);        if(!cnt.count(r))            cnt[r] = 0;          //判断是否出现过不符合条件的单词        cnt[r]++;    }    vector<string> ans;    for(int i = 0; i < words.size(); i++)        if(cnt[re(words[i])] == 1)             ans.push_back(words[i]);       //将符合条件的放入不定长数组vector容器内，注意输入的是words内的含大写的单词    sort(ans.begin(),ans.end());            //排序    for(int i = 0; i < ans.size(); i++)        cout<<ans[i]<<endl;    return 0;}