UVALive 4670 Dominating Patterns

Dominating Patterns

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 4670
64-bit integer IO format: %lld Java class name: Main

The archaeologists are going to decipher a very mysterious ``language". Now, they know many language patterns; each pattern can be treated as a string on English letters (only lower case). As a sub string, these patterns may appear more than one times in a large text string (also only lower case English letters).

What matters most is that which patterns are the dominating patterns. Dominating pattern is the pattern whose appearing times is not less than other patterns.

It is your job to find the dominating pattern(s) and their appearing times.

Input

The entire input contains multi cases. The first line of each case is an integer, which is the number of patterns N, 1N150. Each of the following N lines contains one pattern, whose length is in range [1, 70]. The rest of the case is one line contains a large string as the text to lookup, whose length is up to 106.

At the end of the input file, number `0' indicates the end of input file.

Output

For each of the input cases, output the appearing times of the dominating pattern(s). If there are more than one dominating pattern, output them in separate lines; and keep their input order to the output.

Sample Input

2 
aba 
bab 
ababababac 
6 
beta 
alpha 
haha 
delta 
dede 
tata 
dedeltalphahahahototatalpha 
0

Sample Output

4 
aba 
2 
alpha 
haha

Source

Regionals 2009, Asia - Hefei

解题：Trie图或者AC自动机

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 500;
 4 char str[maxn][maxn];
 5 map<string,int>mp;
 6 struct Trie{
 7     int ch[maxn*maxn][26],fail[maxn*maxn],cnt[maxn*maxn],tot,ret;
 8     bool flag[maxn*maxn];
 9     void init(){
10         ret = tot = 0;
11         newnode();
12     }
13     int newnode(){
14         memset(ch[tot],0,sizeof ch[tot]);
15         fail[tot] = cnt[tot] = 0;
16         flag[tot] = false;
17         return tot++;
18     }
19     void insert(char *str,int root = 0){
20         for(int i = 0; str[i]; ++i){
21             int &x =  ch[root][str[i]-'a'];
22             if(!x) x = newnode();
23             root = x;
24         }
25         mp[str] = root;
26         flag[root] = true;
27     }
28     void build(int rt = 0){
29         queue<int>q;
30         for(int i = 0; i < 26; ++i)
31             if(ch[rt][i]) q.push(ch[rt][i]);
32         while(!q.empty()){
33             rt = q.front();
34             q.pop();
35             for(int i = 0; i < 26; ++i){
36                 int &x = ch[rt][i],y = ch[fail[rt]][i];
37                 if(x){
38                     fail[x] = y;
39                     q.push(x);
40                 }else x = y;
41             }
42         }
43     }
44     void count(char *str,int rt = 0){
45         for(int i = 0; str[i]; ++i){
46             int x = rt = ch[rt][str[i]-'a'];
47             while(x && flag[x]){
48                 ret = max(ret,++cnt[x]);
49                 x = fail[x];
50             }
51         }
52     }
53 }ac;
54 char ss[1000100];
55 int main(){
56     int n;
57     while(scanf("%d",&n),n){
58         ac.init();
59         mp.clear();
60         for(int i = 0; i < n; ++i){
61             scanf("%s",str[i]);
62             ac.insert(str[i]);
63         }
64         ac.build();
65         scanf("%s",ss);
66         ac.count(ss);
67         printf("%d
",ac.ret);
68         for(int i = 0; i < n; ++i)
69             if(ac.cnt[mp[str[i]]] == ac.ret) puts(str[i]);
70     }
71     return 0;
72 }

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