Description
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, 1N
150. 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
AC自动机模版题，注意给的数据末位有一个空格…freopen的时候小心了……
1 #include <algorithm> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <iostream> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <climits> 12 #include <cmath> 13 14 using namespace std; 15 16 inline int GetId(char c) { 17 return c - 'a'; 18 } 19 20 char str[1000010]; 21 char ss[200][110]; 22 int cnt[1000]; 23 int n; 24 int ans; 25 26 typedef class Node { 27 public: 28 Node *next[26]; 29 Node *fail; 30 int id; 31 int cnt; 32 Node(){ 33 for(int i = 0; i < 26; i++) { 34 next[i] = NULL; 35 } 36 fail = NULL; 37 id = -1; 38 cnt = 0; 39 } 40 }Node; 41 42 class AC_Automation { 43 public: 44 Node *root; 45 queue <Node*> q; 46 AC_Automation() { 47 root = new Node; 48 while(!q.empty()) { 49 q.pop(); 50 } 51 } 52 void insert(char *s, int th) { 53 Node *cur = root; 54 int idx; 55 for(int i = 0; s[i]; i++) { 56 idx = GetId(s[i]); 57 if(!cur->next[idx]) { 58 cur->next[idx] = new Node(); 59 } 60 cur = cur->next[idx]; 61 } 62 cur->id = th; 63 cur->cnt++; 64 } 65 void BuildAC() { 66 Node *cur,*tmp; 67 q.push(root); 68 while(!q.empty()) { 69 cur = q.front(); 70 q.pop(); 71 for(int i = 0; i < 26; i++) { 72 if(cur->next[i]) { 73 if(cur == root) { 74 cur->next[i]->fail = root; 75 } 76 else { 77 tmp = cur->fail; 78 while(tmp->fail && !tmp->next[i]) { 79 tmp = tmp->fail; 80 } 81 if(tmp->next[i]) { 82 cur->next[i]->fail = tmp->next[i]; 83 } 84 else { 85 cur->next[i]->fail = root; 86 } 87 } 88 q.push(cur->next[i]); 89 } 90 } 91 } 92 } 93 void query(char *s) { 94 Node *p = root; 95 Node *tmp; 96 char x; 97 for(int i = 0; s[i]; i++) { 98 x = GetId(s[i]); 99 while(!p->next[x] && p != root) { 100 p = p->fail; 101 } 102 p = p->next[x]; 103 if(!p) { 104 p = root; 105 } 106 tmp = p; 107 while(tmp != root) { 108 if(tmp->cnt >= 1) { 109 if(tmp->id != -1) { 110 cnt[tmp->id]++; 111 } 112 } 113 tmp = tmp->fail; 114 } 115 } 116 } 117 }; 118 119 int main() { 120 // freopen("in", "r", stdin); 121 while(~scanf("%d", &n) && n) { 122 memset(cnt, 0, sizeof(cnt)); 123 memset(str, 0, sizeof(str)); 124 memset(ss, 0, sizeof(ss)); 125 getchar(); 126 AC_Automation ac; 127 ans = -1; 128 for(int i = 0; i < n; i++) { 129 gets(ss[i]); 130 ac.insert(ss[i], i); 131 } 132 ac.BuildAC(); 133 gets(str); 134 ac.query(str); 135 for(int i = 0; i < n; i++) { 136 if(cnt[i] > ans) { 137 ans = cnt[i]; 138 } 139 } 140 printf("%d ", ans); 141 for(int i = 0; i < n; i++) { 142 if(ans == cnt[i]) { 143 printf("%s ", ss[i]); 144 } 145 } 146 } 147 }