字符串的長(zhǎng)度有限(最多只有12位)，我建一個(gè)trie樹(shù)來(lái)存儲(chǔ)所有字符串，然后遍歷trie樹(shù)，將對(duì)應(yīng)的字符串存儲(chǔ)到一個(gè)vector中，再排序輸出。
ac以后，看usaco的分析，是用位串來(lái)做索引計(jì)數(shù)，這種方法比較簡(jiǎn)潔巧妙。為了解決前綴為0的問(wèn)題，在每個(gè)字符串前面加了一個(gè)1，輸出的時(shí)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;
}
附題：
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