题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2222

题意：给你一系列子串，再给你一个主串问你主串一共有几个匹配子串

原来使用字典树写的但数据有点大TLE了，然后就开始学习ac自动机了，ac自动机就像是多串匹配的kmp原理也是类似的。

这题可以练一下手写ac自动机模版。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int M = 5e5 + 50;
int Next[M][26] , fail[M] , End[M] , root , L , cnt;
char str[60] , sl[M * 2];
int newnode() {for(int i = 0 ; i < 26 ; i++) {Next[L][i] = -1;}End[L++] = 0;return L - 1;
}
void init() {L = 0;root = newnode();
}
void build(char s[]) {int len = strlen(s);int now = root;for(int i = 0 ; i < len ; i++) {int id = s[i] - 'a';if(Next[now][id] == -1) {Next[now][id] = newnode();}now = Next[now][id];}++End[now];
}
void get_fail() {queue<int>q;fail[root] = root;for(int i = 0 ; i < 26 ; i++) {if(Next[root][i] == -1) {Next[root][i] = root;}else {fail[Next[root][i]] = root;q.push(Next[root][i]);}}while(!q.empty()) {int now = q.front();q.pop();for(int i = 0 ; i < 26 ; i++) {if(Next[now][i] == -1) {Next[now][i] = Next[fail[now]][i];}else {fail[Next[now][i]] = Next[fail[now]][i];q.push(Next[now][i]);}}}
}
void match(char s[]) {int now = root;int len = strlen(s);for(int i = 0 ; i < len ; i++) {int id = s[i] - 'a';now = Next[now][id];int temp = now;while(temp != root) {cnt += End[temp];End[temp] = 0;temp = fail[temp];}}
}
int main()
{int t;scanf("%d" , &t);while(t--) {int n;scanf("%d" , &n);init();for(int i = 0 ; i < n ; i++) {scanf("%s" , str);build(str);}get_fail();scanf("%s" , sl);cnt = 0;match(sl);printf("%d\n" , cnt);}return 0;
}