原题： 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 中文： 题目中定义了一个重复的类型，如果字符串a与字符串b不分大小写而且a经过字母重排后能变成b。那么就算a与b字母重复。现在输入一个文本，让你把出现重复的单词全去掉，留下没出现重复过的单词，然后按照（原字符串）字典序排序后输出。
#include <bits/stdc++.h> using namespace std; map<string,string> ms; string s; vector<string> vs; string get_lower_sort(string ss) { for(auto &x:ss) x=tolower(x); sort(ss.begin(),ss.end()); return ss; } int main() { ios::sync_with_stdio(false); while(cin>>s) { if(s=="#") { for(auto x:ms) if(x.second!="#") vs.push_back(x.second); sort(vs.begin(),vs.end()); for(auto x:vs) cout<<x<<endl; ms.clear(); vs.clear(); } else { string tmp=get_lower_sort(s); if(ms[tmp].empty()) ms[tmp]=s; else ms[tmp]="#"; } } return 0; }解答： 很简单的一道应用map的题目，首先定义一个函数，把原字符串全部转换成小写，然后再把这个字符里的字母串排序，返回成ss。定义一个map，key值对应ss，value对应原字符串。如果对应的key值出现重复，那么就把这个字符串存为#。全部文本处理完后把map当中value值不为#的值保存到vector中，排序输出即可。
