UVa 156 Ananagrams (用STL multimap&set处理字典中的重复元素)

寻找相对Ananagrams

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本文介绍了一种通过程序判断特定领域内单词是否为相对Ananagrams的方法，即那些无论怎样重新排列字母都无法形成另一个该领域内单词的特殊词汇。文章详细阐述了实现这一功能的具体步骤，并提供了完整的代码示例。

156 - Ananagrams

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=98&page=show_problem&problem=92

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

思路：先用建立一个multimap<排序的词,未排序的词>来处理输入，完毕后将只有一个值的键放入一个set中，最后遍历set输出即可。（详见代码）

完整代码：

/*0.015s*/

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
using namespace std;

char s[25], origin[25];
multimap<string, string> map1;
set<string> set1;
multimap<string, string>::iterator pm;
set<string>::iterator ps;

int main()
{
	int len, i;
	while (scanf("%s", s), s[0] != '#')
	{
		strcpy(origin, s);
		len = strlen(s);
		for (i = 0; i < len; ++i)
			if (isupper(s[i])) s[i] = tolower(s[i]);
		sort(s, s + len);
		map1.insert(make_pair(s, origin));
	}
	for (pm = map1.begin(); pm != map1.end(); ++pm)
		if (map1.count(pm->first) == 1)
			set1.insert(pm->second);
	for (ps = set1.begin(); ps != set1.end(); ++ps)
		puts((*ps).c_str());
	return 0;
}