USACO 3.1 Contact

USACO 3.1 Contact

字符串的长度有限(最多只有12位)，我建一个trie树来存储所有字符串，然后遍历trie树，将对应的字符串存储到一个vector中，再排序输出。
ac以后，看usaco的分析，是用位串来做索引计数，这种方法比较简洁巧妙。为了解决前缀为0的问题，在每个字符串前面加了一个1，输出的时候再去掉。
我的解法如下：

#include  < iostream >
#include  < fstream >
#include  < vector >
#include  < string >

using   namespace  std;

ifstream fin( " contact.in " );
ofstream fout( " contact.out " );

#ifdef _DEBUG
#define  out cout
#define  in cin
#else
#define  out fout
#define  in fin
#endif

struct  trie_node{
     int  cnt;
    trie_node *  sons[ 2 ];
    trie_node(){
        sons[ 0 ]  =  sons[ 1 ]  =   0 ;
        cnt  =   0 ;
    }
};

struct  sort_node{
     string  str;
     int  cnt;

     bool   operator < ( const  sort_node & n2)  const {
         if (cnt != n2.cnt)  return  cnt > n2.cnt;
         if (str.size() != n2.str.size())  return  str.size() < n2.str.size();
         for ( int  i = 0 ;i < str.size(); ++ i){
             if (str[i] != n2.str[i])
                 return  str[i] < n2.str[i];
        }
    }
};

char  buf[ 200000 ];
int  buf_len;
int  a,b,n;
vector < sort_node > res;

void  insert_trie(trie_node * root, const   char   * start, int  len)
{
    trie_node  *  next  =  root;

     for ( int  i = 0 ;i < len; ++ i){
         if (next -> sons[start[i] - ' 0 ' ] == NULL)
            next -> sons[start[i] - ' 0 ' ]  =   new  trie_node;
        next  =  next -> sons[start[i] - ' 0 ' ];
         if (i + 1 >= a && i + 1 <= b)
            next -> cnt ++ ;
    }
}

void  _traverse_trie(trie_node  * root, char  ch, int  depth)
{
     if (root == NULL)  return ;

    buf[depth] = ch;

     if (root -> cnt != 0 ){
        sort_node n;
        n.str  =   string ( & buf[ 0 ], & buf[depth + 1 ]);
        n.cnt  =  root -> cnt;
        res.push_back(n);
    }

    _traverse_trie(root -> sons[ 0 ], ' 0 ' ,depth + 1 );
    _traverse_trie(root -> sons[ 1 ], ' 1 ' ,depth + 1 );
}

void  traverse_trie(trie_node  * root)
{
    _traverse_trie(root -> sons[ 0 ], ' 0 ' , 0 );
    _traverse_trie(root -> sons[ 1 ], ' 1 ' , 0 );
}


void  solve()
{
     in >> a >> b >> n;

     char  ch;
     while ( in . get (ch)){
         if (ch == ' 0 ' || ch == ' 1 ' )
            buf[buf_len ++ ]  =  ch;
    }

    trie_node root;

     for ( int  i = 0 ;i + b <= buf_len; ++ i){
        insert_trie( & root, & buf[i],b);
    }

     for ( int  i = min(b - 1 ,buf_len);i >= a; -- i){
        insert_trie( & root, & buf[buf_len - i],i);
    }

    traverse_trie( & root);

    sort(res.begin(),res.end());

     int  freq_cnt  =   0 ;
     int  last_cnt  =   - 1 ;
     int  out_cnt  =   0 ;

     for ( int  i = 0 ;i < res.size(); ++ i){
         if (res[i].cnt == last_cnt){
             if (out_cnt % 6 != 0 )
             out << "   " ;
             out << res[i].str;
            out_cnt ++ ;
             if (out_cnt % 6 == 0 || i == (res.size() - 1 ) || res[i + 1 ].cnt != res[i].cnt)
                 out << endl;
        } else {
            last_cnt  =  res[i].cnt;
            freq_cnt ++ ;
             if (freq_cnt > n)
                break ; 
             out << res[i].cnt << endl;
             out << res[i].str;
            out_cnt  =   0 ;
            out_cnt ++ ;
             if ( (i == res.size() - 1 ) || res[i + 1 ].cnt != res[i].cnt)
                 out << endl;
        }
    }
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}

附题：
Contact
IOI'98

The cows have developed a new interest in scanning the universe outside their farm with radiotelescopes. Recently, they noticed a very curious microwave pulsing emission sent right from the centre of the galaxy. They wish to know if the emission is transmitted by some extraterrestrial form of intelligent life or if it is nothing but the usual heartbeat of the stars.

Help the cows to find the Truth by providing a tool to analyze bit patterns in the files they record. They are seeking bit patterns of length A through B inclusive (1 <= A <= B <= 12) that repeat themselves most often in each day's data file. They are looking for the patterns that repeat themselves most often. An input limit tells how many of the most frequent patterns to output.

Pattern occurrences may overlap, and only patterns that occur at least once are taken into account.

PROGRAM NAME: contact

INPUT FORMAT

Line 1:	Three space-separated integers: A, B, N; (1 <= N < 50)
Lines 2 and beyond:	A sequence of as many as 200,000 characters, all 0 or 1; the characters are presented 80 per line, except potentially the last line.

SAMPLE INPUT (file contact.in)

2 4 10
01010010010001000111101100001010011001111000010010011110010000000

In this example, pattern 100 occurs 12 times, and pattern 1000 occurs 5 times. The most frequent pattern is 00, with 23 occurrences.

OUTPUT FORMAT

Lines that list the N highest frequencies (in descending order of frequency) along with the patterns that occur in those frequencies. Order those patterns by shortest-to-longest and increasing binary number for those of the same frequency. If fewer than N highest frequencies are available, print only those that are.

Print the frequency alone by itself on a line. Then print the actual patterns space separated, six to a line (unless fewer than six remain).

SAMPLE OUTPUT (file contact.out)

23
00
15
01 10
12
100
11
11 000 001
10
010
8
0100
7
0010 1001
6
111 0000
5
011 110 1000
4
0001 0011 1100