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            Brian Warehouse

            Some birds aren`t meant to be caged, their feathers are just too bright... ...
            posts - 40, comments - 16, trackbacks - 0, articles - 1

            POJ 1002 487-3279

            Posted on 2010-08-17 14:04 Brian 閱讀(259) 評(píng)論(0)  編輯 收藏 引用 所屬分類(lèi): POJ

            事先申明,該程序雖然AC,但是效率極其低下,低下到讓人發(fā)指的程度,我也不知道為什么。估計(jì)是用了STL的原因,具體我也說(shuō)不清楚。其實(shí)思路不難,就是將字符轉(zhuǎn)化成對(duì)應(yīng)數(shù)字,然后將結(jié)果存放在一個(gè)整型向量中,接收字符串用的是字符串向量,處理的時(shí)候跟一般的字符串處理時(shí)一模一樣的。處理結(jié)束之后要進(jìn)行字典排序,顯然要用排序函數(shù),可以用冒泡,選擇,快排,甚至是Hash,但是據(jù)說(shuō)STL的sort 效率比快排還要快。源程序后附加了MSDN上的一些簡(jiǎn)單解釋。沒(méi)有翻譯!

            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.

            C++ 編譯器:

            #include <iostream>
            #include <string>
            #include <vector>
            #include <algorithm> // STL sort function
            using namespace std;

            char map[] = "2223334445556667#77888999#";
                              //ABCDEFGHIJKLMNOPQRSTUVWXYZ
            void visited(char &ch) // visit and format strings
            {
               if (ch >= 'A' && ch <= 'Z')
                ch=map[ch-'A']; // ch equals to its real number
            }

            int main()
            {
                int N,i=0,j,flag=0;
             string s;
             vector<string> stored(100000); // be visited & stored (up to 100,000)
                cin>>N;
             vector<int> counter(N,1); // stored times
               
             for (; i<N; i++)
             {
              cin>>s;
              for (j=0; j<s.length(); j++) // MSDN
              {
               visited(s[j]);
               if (s[j]!='-')
               {
                stored[i] += s[j];
                if (stored[i].length()==3)
                 stored[i] += '-'; // 487 -[3] 3279
               }
              }
             }
                sort(stored.begin(),stored.begin()+N); // Quicker than QuickSort!
             // should not used stored.end() !
             i=0; j=1;
                while (i<N)
             {
              while(stored[i] == stored[j])
              {
               counter[i]++;
               j++;
               flag=1;
              }
              i=j;
              j++;
             }
             
                if (flag)
              for (i=0; i<N; i++)
              {
               if (counter[i]>1)
                cout<<stored[i]<<" "<<counter[i]<<endl;
              } // must have { }
             else cout<<"No duplicates."<<endl;
              
             return 0;
            }

            Sort :
            Arranges the elements in a specified range into a nondescending order or according to an ordering criterion specified by a binary predicate.

             
            template<class RandomAccessIterator>
               void sort(
                  RandomAccessIterator _First,
                  RandomAccessIterator _Last
               );

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