UVA-156 Ananagrams

最新推荐文章于 2021-02-01 19:31:28 发布

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最新推荐文章于 2021-02-01 19:31:28 发布

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本文介绍了一种程序设计方法，用于识别特定领域的相对Ananagrams（即无法通过字母重组形成同一领域内其他词汇的单词）。文章详细解释了相对Ananagrams的概念，并提供了具体的实现方案，包括如何标准化单词、使用map容器进行存储以及最终筛选并输出结果。

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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  drIed
noel dire Disk mace Rob dries
#

Sample output

Disk
NotE
derail
drIed
eye
ladder
soon

_____________________________________________________________________________

题目大意：输入一段话然后输出重排后没有重复的单词。

这道题紫书上面有看了之后按照自己的思路写了一遍，大概思路就是先把每一个单词标准化（就是不分大小写，按字典序排列），然后通过map容器存下来，如过map对应的值为一，就证明已经有了，就把map的值变为零，然后遍历一下map为一的元素，存到vector里面，然后排序输出就行了

有个需要注意的技巧，用三个数组来保存输入时的单词，标准化的单词，以及要输出的单词（这里不能去确定数组的大小所以选择用vector）

下面是ac代码

#include<cstdio>
#include<map>
#include<vector>
#include<algorithm>
#include<string.h>
#include<iostream>
using namespace std;
map<string,int>q;
vector<string>s,ss;
string bzh(string sss)//单词的标准化
{
int len=sss.length();
for(int i=0;i<len;i++)
{
if(sss[i]>='A'&&sss[i]<='Z')
sss[i]+=32;

}
sort(sss.begin(),sss.end());
return sss;
}
int main()
{
string s1;
while(cin>>s1)
{
if(s1[0]=='#')
break;
s.push_back(s1);
s1=bzh(s1);
if(!q.count(s1))
q[s1]=0;
q[s1]++;
}
for(int i=0;i<s.size();i++)
{
if(q[bzh(s[i])]==1)
{
ss.push_back(s[i]);
}
}
sort(ss.begin(),ss.end());
for(int i=0;i<ss.size();i++)
cout<<ss[i]<<endl;
return 0;
}