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'98The 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