POJ_1002 487-3279 题解

我就是我 2021-11-19 16:48 265阅读 0赞

**一、题目描述**

**487-3279**

<table> 
 <tbody> 
  <tr> 
   <td><span style="font-size:12px;"><strong>Time Limit:</strong>&nbsp;2000MS</span></td> 
   <td>&nbsp;</td> 
   <td><span style="font-size:12px;"><strong>Memory Limit:</strong>&nbsp;65536K</span></td> 
  </tr> 
  <tr> 
   <td><span style="font-size:12px;"><strong>Total Submissions:</strong>&nbsp;180121</span></td> 
   <td>&nbsp;</td> 
   <td><span style="font-size:12px;"><strong>Accepted:</strong>&nbsp;31015</span></td> 
  </tr> 
 </tbody> 
</table>

**Description**

Businesses like to have memorable telephone numbers. One way to make a telephone number memorable is to have it spell a memorable word or phrase. For example, you can call the University of Waterloo by dialing the memorable TUT-GLOP. Sometimes only part of the number is used to spell a word. When you get back to your hotel tonight you can order a pizza from Gino's by dialing 310-GINO. Another way to make a telephone number memorable is to group the digits in a memorable way. You could order your pizza from Pizza Hut by calling their \`\`three tens'' number 3-10-10-10.   
The standard form of a telephone number is seven decimal digits with a hyphen between the third and fourth digits (e.g. 888-1200). The keypad of a phone supplies the mapping of letters to numbers, as follows:   
A, B, and C map to 2   
D, E, and F map to 3   
G, H, and I map to 4   
J, K, and L map to 5   
M, N, and O map to 6   
P, R, and S map to 7   
T, U, and V map to 8   
W, X, and Y map to 9   
There is no mapping for Q or Z. Hyphens are not dialed, and can be added and removed as necessary. The standard form of TUT-GLOP is 888-4567, the standard form of 310-GINO is 310-4466, and the standard form of 3-10-10-10 is 310-1010.   
Two telephone numbers are equivalent if they have the same standard form. (They dial the same number.)   
Your company is compiling a directory of telephone numbers from local businesses. As part of the quality control process you want to check that no two (or more) businesses in the directory have the same telephone number.

**Input**

The input will consist of one case. The first line of the input specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters.

**Output**

Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line:   
No duplicates.

**Sample Input**

12
    4873279
    ITS-EASY
    888-4567
    3-10-10-10
    888-GLOP
    TUT-GLOP
    967-11-11
    310-GINO
    F101010
    888-1200
    -4-8-7-3-2-7-9-
    487-3279

**Sample Output**

310-1010 2
    487-3279 4
    888-4567 3

二、解题思路

一道让我忙活了半天的题目，开始先想到将字符串转换成整型数，然后对整型数的大小排序即是按照字符串的字典顺序排序，因此步骤有三：

（1）将字符串根据题目要求全部转换成整型数；

（2）快速排序；

（3）按照输出格式要求输出重复数。

但是不知道为什么，这么写总是Time Limitation Exceeds，需要进一步思考。

后来想到利用C++标准库中的<algorithm>中的sort()函数，直接对string类型进行排序，结果AC了。

讨论：

（1）字符串转换成整型数以及将整型数转换成字符串需损耗一定时间，无需转换；

（2）善于利用C++标准库，有时根据需要对运算符进行重载；

（3）在网上发现此题还有其它如HASH或其它算法，需要进一步讨论。

三、源代码

1 #include <iostream>
      2 #include <string>
      3 #include <vector>
      4 #include <algorithm>
      5 
      6 using namespace std;
      7 
      8 string convert_str(string &str)
      9 {
        
     10     string s = "";
     11     for (int i = 0; i < str.length(); i++)
     12     {
        
     13         if (str[i] == 'A' || str[i] == 'B' || str[i] == 'C')
     14         {
        
     15             s.push_back('2');
     16         }
     17         else if (str[i] == 'D' || str[i] == 'E' || str[i] == 'F')
     18         {
        
     19             s.push_back('3');
     20         }
     21         else if (str[i] == 'G' || str[i] == 'H' || str[i] == 'I')
     22         {
        
     23             s.push_back('4');
     24         }
     25         else if (str[i] == 'J' || str[i] == 'K' || str[i] == 'L')
     26         {
        
     27             s.push_back('5');
     28         }
     29         else if (str[i] == 'M' || str[i] == 'N' || str[i] == 'O')
     30         {
        
     31             s.push_back('6');
     32         }
     33         else if (str[i] == 'P' || str[i] == 'R' || str[i] == 'S')
     34         {
        
     35             s.push_back('7');
     36         }
     37         else if (str[i] == 'T' || str[i] == 'U' || str[i] == 'V')
     38         {
        
     39             s.push_back('8');
     40         }
     41         else if (str[i] == 'W' || str[i] == 'X' || str[i] == 'Y')
     42         {
        
     43             s.push_back('9');
     44         }
     45         else if (str[i] == '-')
     46         {
        
     47             continue;
     48         }
     49         else
     50         {
        
     51             s.push_back(str[i]);
     52         }
     53     }
     54     return s;
     55 }
     56 
     57 void convert_str2(string &str)
     58 {
        
     59     str.insert(3,"-");
     60 }
     61 
     62 
     63 int main()
     64 {
        
     65     int n;
     66     cin >> n;
     67     vector<string> telenum;
     68     for (int i=0; i < n; i++)
     69     {
        
     70         string tmp = "";
     71         cin >> tmp;
     72         telenum.push_back(convert_str(tmp));
     73     }
     74 
     75     sort(telenum.begin(),telenum.end());
     76 
     77     int index = 0;
     78     int count = 1;
     79     int flag = 0;
     80     for (int i = 1; i < n; i++)
     81     {
        
     82         if (telenum[i] == telenum[index])
     83         {
        
     84             count++;
     85             flag = 1;
     86         }
     87         else
     88         {
        
     89             if (count == 1)
     90             {
        
     91                 index = i;
     92             }
     93             else
     94             {
        
     95                 flag = 2;
     96                 convert_str2(telenum[index]);
     97                 cout << telenum[index] << " " << count << endl;
     98                 index = i;
     99                 count = 1;
    100             }
    101         }
    102     }
    103 
    104     if (flag == 0)
    105     {
        
    106         cout << "No duplicates." << endl;
    107     }
    108     else if (flag == 1)
    109     {
        
    110         convert_str2(telenum[index]);
    111         cout << telenum[index] << " " << count << endl;
    112     }
    113     return 0;
    114 }

转载于:https://www.cnblogs.com/caoxiao/archive/2012/03/02/2377331.html