PKU 2408 Anagram Groups (排序)

PKU 2408 Anagram Groups (排序)

Anagram Groups

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2318		Accepted: 649

Description

World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups. 

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group. Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace(or new line). It is terminated by EOF. You can assume there will be no more than 30000 words.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

Source

Ulm Local 2000

思路:
这题将排序发挥到了极致啊呵呵，排序来排序去就AC了

代码:

 1  /*  47MS  */
 2  #include < stdio.h >
 3  #include < stdlib.h >
 4  #include < string .h >
 5  #define  MAX_NUM 30001
 6  #define  MAX_LEN 36
 7  #define  MAX_OUT 5
 8  struct  Word {
 9       char  word[MAX_LEN];
10       char  word_cmp[MAX_LEN];
11  } words[MAX_NUM];
12 
13  struct  Summary {
14       struct  Word  * first;
15       int  count;
16  } smmry[MAX_NUM];
17 
18  int  total, total_category;
19 
20  int
21  cmp_char( const   void   * arg1,  const   void   * arg2)
22  {
23       return  ( * ( char   * )arg1)  -  ( * ( char   * )arg2);
24  }
25 
26  int
27  cmp_words( const   void   * arg1,  const   void   * arg2)
28  {
29       int  ret  =  strcmp((( struct  Word  * )arg1) -> word_cmp, (( struct  Word  * )arg2) -> word_cmp);
30       if (ret  ==   0 )
31          ret  =  strcmp((( struct  Word  * )arg1) -> word, (( struct  Word  * )arg2) -> word);
32       return  ret;
33  }
34 
35  int
36  cmp_category( const   void   * arg1,  const   void   * arg2)
37  {
38       int  ret  =  (( struct  Summary  * )arg2) -> count  -  (( struct  Summary  * )arg1) -> count;
39       if (ret  ==   0 )
40          ret  =  strcmp((( struct  Summary  * )arg1) -> first -> word, (( struct  Summary  * )arg2) -> first -> word);
41       return  ret;
42  }
43 
44  int
45  main( int  argc,  char   ** argv)
46  {
47       int  i, j, num, len;
48      total  =  total_category  =   0 ;
49       while (scanf( " %s " , words[total].word)  !=  EOF) {
50          len  =  strlen(words[total].word);
51          strcpy(words[total].word_cmp, words[total].word);
52          qsort(words[total].word_cmp, len,  sizeof ( char ), cmp_char); 
53           ++ total;
54      }
55      qsort(words, total,  sizeof ( struct  Word), cmp_words);
56 
57      num  =   1 ;
58       for (i = 1 ; i < total; i ++ ) {
59           if (strcmp(words[i].word_cmp, words[i - 1 ].word_cmp)  ==   0 )
60               ++ num;
61           else  {
62              smmry[total_category].first  =  words + i - num;
63              smmry[total_category].count  =  num;
64               ++ total_category;
65              num  =   1 ;
66          }
67      }
68      smmry[total_category].first  =  words + i - num;
69      smmry[total_category ++ ].count  =  num;
70      qsort(smmry, total_category,  sizeof ( struct  Summary), cmp_category);
71 
72      total_category  =  total_category  <  MAX_OUT  ?  total_category : MAX_OUT;
73       for (i = 0 ; i < total_category; i ++ ) {
74          printf( " Group of size %d: %s  " , smmry[i].count, smmry[i].first -> word);
75           for (j = 1 ; j < smmry[i].count; j ++ )
76               if (strcmp((smmry[i].first + j) -> word, (smmry[i].first + j - 1 ) -> word)  !=   0 )
77                  printf( " %s  " , (smmry[i].first + j) -> word);
78          printf( " .\n " );
79      }
80  }