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    Description

    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
    解题思路:题目的意思就是找出唯一的(不与其他字符串串内重排后相同的)字符串,将它们排序并输出。做法:用map记录所有字符串内部字符(所有大写改成小写。为什么呢?上面的红色部分就已经解释了,我们需要统一换成大写或者小写)重排后的字符串的个数,用两个vector,一个用来保存文本中的所有单词,另一个保存出现次数为1的单词,最后排序输出即可。
    AC代码:
     1 #include<bits/stdtr1c++.h>
     2 using namespace std;
     3 string repeat(string tmp){
     4     for(size_t i=0;i<tmp.size();++i)
     5         if(isupper(tmp[i]))tmp[i]+=32;//如果是大写字母,全部改为小写字母
     6     sort(tmp.begin(),tmp.end());//串内重新排序
     7     return tmp;
     8 }
     9 int main(){
    10     string str,tp;
    11     vector<string> vec,vrc;
    12     map<string,int> mp;//记录串内字符重排后出现的次数
    13     while(cin>>str&&str!="#"){
    14         vec.push_back(str);
    15         tp=repeat(str);
    16         if(!mp.count(tp))mp[tp]=0;//如果还没有出现,次数先初始化为0
    17         mp[tp]++;//统计相同字符串出现的次数
    18     }
    19     vrc.clear();//清空容器
    20     for(size_t i=0;i<vec.size();++i)
    21         if(mp[repeat(vec[i])]==1)vrc.push_back(vec[i]);//如果该单词出现的次数为1,则满足条件,保存在vrc容器中
    22     sort(vrc.begin(),vrc.end());//排序
    23     for(size_t i=0;i<vrc.size();++i)
    24         cout<<vrc[i]<<endl;
    25     return 0;
    26 }
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  • 原文地址:https://www.cnblogs.com/acgoto/p/9413952.html
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