UVA156_map

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

题意：

输入一些单词，找出所有满足以下条件的：该单词不能通过字母重排，得到输入文本中的另外一个单词。在判断是否满足条件时，不区分字母大小写，但在输出时应保留输入中的大小写，按字典序进行排列（所有大写字母在所有小写字母前面）。

#include <iostream>
#include <vector>
#include <string>
#include <map>
#include <cctype>
#include <algorithm>
using namespace std;
map<string,int> cnt;
vector<string> words;
string repr(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;
}
int main()
{
    string s;
    while(cin>>s&&s!="#")
    {
        words.push_back(s);
        string r=repr(s);
        if(!cnt.count(r))   cnt[r]=0;	//cnt中存在r则返回1，不存在返回0；
        cnt[r]++;
    }
    vector<string> ans;
    for(int i=0;i<words.size();i++)
        if(cnt[repr(words[i])]==1)  ans.push_back(words[i]);
    sort(ans.begin(),ans.end());
    for(int i=0;i<ans.size();i++)
    cout<<ans[i]<<endl;
    return 0;
}