UVA - 156 Ananagrams ——STL练习

你的名字 2022-05-19 06:50 149阅读 0赞

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：  
输入一些单词，按字典序顺序输出**无视大小**写及**字母顺序**后，只出现一次的单词。如 STOP、SOPT、spTO 算重复出现。

--------------------

#include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <string>
    #include <map>
    #include <set>
    #include <vector>
    #include <cstring>
    using namespace std;
    
    string change(string s)
    {
        
        for(int i=0;i<s.size();i++)
            s[i]=tolower(s[i]);
        sort(s.begin(),s.end());//按字典序排列单词
        return s;
    }
    int main()
    {
        
        string s;
        map<string ,int > p;//用来统计单词 change 后的次数
        vector<string> ans;//用来保存结果
        vector<string> ve;//用来保存原始单词，因为输出的时候就是原始单词
        while(cin>>s&&s!="#")
        {
        
            ve.push_back(s);
            p[change(s)]++;
        }
        for(int i=0;i<ve.size();i++)
            if(p[change(ve[i])]==1)//如果单词改变后只出现了一次，那就符合答案
                ans.push_back(ve[i]);
        sort(ans.begin(),ans.end());//排序，按字典序升序排列
        for(int i=0;i<ans.size();i++)
            cout<<ans[i]<<endl;
        return 0;
    }

或者

#include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const ll MAXN = 2e5+7;
    // ll a[MAXN];
    struct node{
        
            int num;
            string s;
            node(){
        num=0;};//在结构体内部初始化值
            node(int _num, string _s)
            {
        
                    num = _num;
                    s = _s;
            }
    };
    int main()
    {
        
            set<string> se;
            map<string,node> m;
            string str;
            while(cin>>str)
            {
        
                    string c;
                    if(str=="#")    break;
                    for(int i=0;i<str.size();++i)
                            c += tolower(str[i]);
                    sort(c.begin(),c.end());
                    m[c] = node(m[c].num+1,str);
            }
    
            for(map<string,node>::iterator it=m.begin();it!=m.end();++it)
                    if(it->second.num==1)   se.insert(it->second.s);
            for(set<string>::iterator it=se.begin();it!=se.end();it++)
                    cout<<*it<<endl;
            return 0;
    }