86 lines
2.3 KiB
C++
86 lines
2.3 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <algorithm>
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#include <iostream>
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using namespace std;
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const int N = 1000010;
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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
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char T[N]; //文本串 ，长度最大10^6
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int n; //模式串数量
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int cnt[N]; //每个模式串出现的次数
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int tr[N][26], idx; // Trie树
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int id[N]; // 节点号-mapping->模式串
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void insert(char *s, int x) {
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int p = 0;
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for (int i = 0; s[i]; i++) {
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int t = s[i] - 'a';
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if (!tr[p][t]) tr[p][t] = ++idx;
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p = tr[p][t];
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}
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id[p] = x; //记录:节点号-mapping->模式串
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}
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//构建AC自动机
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int q[N], ne[N];
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void bfs() {
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int hh = 0, tt = -1;
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for (int i = 0; i < 26; i++)
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if (tr[0][i]) q[++tt] = tr[0][i];
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while (hh <= tt) {
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int p = q[hh++];
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for (int i = 0; i < 26; i++) {
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int t = tr[p][i];
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if (!t)
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tr[p][i] = tr[ne[p]][i];
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else {
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ne[t] = tr[ne[p]][i];
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q[++tt] = t;
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}
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}
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}
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}
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//查询字符串s在AC自动机中出现的次数
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void query(char *s) {
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int p = 0;
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for (int i = 0; s[i]; i++) {
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p = tr[p][s[i] - 'a'];
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for (int j = p; j; j = ne[j])
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if (id[j]) cnt[id[j]]++; //如果有模式串标记，更新出现次数
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}
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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while (cin >> n && n) {
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//每次清空
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memset(tr, 0, sizeof tr);
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memset(cnt, 0, sizeof cnt);
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memset(ne, 0, sizeof ne);
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memset(id, 0, sizeof id);
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idx = 0;
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for (int i = 1; i <= n; i++) {
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cin >> s[i];
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insert(s[i], i);
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}
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bfs();
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cin >> T;
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query(T);
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int Max = 0;
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for (int i = 1; i <= n; i++) Max = max(cnt[i], Max); //最后统计答案
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printf("%d\n", Max);
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//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
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for (int i = 1; i <= n; i++)
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if (cnt[i] == Max) printf("%s\n", s[i]);
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}
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return 0;
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} |