|
|
#include <cstdio>
|
|
|
#include <cstring>
|
|
|
#include <algorithm>
|
|
|
#include <iostream>
|
|
|
using namespace std;
|
|
|
|
|
|
const int N = 1000010;
|
|
|
char s[150 + 10][70 + 10]; //模式串,第一维是多少个,第二维是具体的字符
|
|
|
char T[N]; //文本串 ,长度最大10^6
|
|
|
int n; //模式串数量
|
|
|
int cnt[N]; //每个模式串出现的次数
|
|
|
|
|
|
int tr[N][26], idx; // Trie树
|
|
|
int id[N]; // 节点号-mapping->模式串
|
|
|
|
|
|
void insert(char *s, int x) {
|
|
|
int p = 0;
|
|
|
for (int i = 0; s[i]; i++) {
|
|
|
int t = s[i] - 'a';
|
|
|
if (!tr[p][t]) tr[p][t] = ++idx;
|
|
|
p = tr[p][t];
|
|
|
}
|
|
|
id[p] = x; //记录:节点号-mapping->模式串
|
|
|
}
|
|
|
|
|
|
//构建AC自动机
|
|
|
int q[N], ne[N];
|
|
|
void bfs() {
|
|
|
int hh = 0, tt = -1;
|
|
|
for (int i = 0; i < 26; i++)
|
|
|
if (tr[0][i]) q[++tt] = tr[0][i];
|
|
|
|
|
|
while (hh <= tt) {
|
|
|
int p = q[hh++];
|
|
|
for (int i = 0; i < 26; i++) {
|
|
|
int t = tr[p][i];
|
|
|
if (!t)
|
|
|
tr[p][i] = tr[ne[p]][i];
|
|
|
else {
|
|
|
ne[t] = tr[ne[p]][i];
|
|
|
q[++tt] = t;
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
|
|
|
//查询字符串s在AC自动机中出现的次数
|
|
|
void query(char *s) {
|
|
|
int p = 0;
|
|
|
for (int i = 0; s[i]; i++) {
|
|
|
p = tr[p][s[i] - 'a'];
|
|
|
for (int j = p; j; j = ne[j])
|
|
|
if (id[j]) cnt[id[j]]++; //如果有模式串标记,更新出现次数
|
|
|
}
|
|
|
}
|
|
|
|
|
|
int main() {
|
|
|
//加快读入
|
|
|
ios::sync_with_stdio(false), cin.tie(0);
|
|
|
while (cin >> n && n) {
|
|
|
//每次清空
|
|
|
memset(tr, 0, sizeof tr);
|
|
|
memset(cnt, 0, sizeof cnt);
|
|
|
memset(ne, 0, sizeof ne);
|
|
|
memset(id, 0, sizeof id);
|
|
|
idx = 0;
|
|
|
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
cin >> s[i];
|
|
|
insert(s[i], i);
|
|
|
}
|
|
|
bfs();
|
|
|
|
|
|
cin >> T;
|
|
|
query(T);
|
|
|
|
|
|
int Max = 0;
|
|
|
for (int i = 1; i <= n; i++) Max = max(cnt[i], Max); //最后统计答案
|
|
|
printf("%d\n", Max);
|
|
|
|
|
|
//最大值可能很多个模式串匹配到,需要在获取完最大值后,再次循环输出符合最大值条件的所有模式串
|
|
|
for (int i = 1; i <= n; i++)
|
|
|
if (cnt[i] == Max) printf("%s\n", s[i]);
|
|
|
}
|
|
|
return 0;
|
|
|
} |