Description:

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input:

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters ‘a’-‘z’, and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

Output:

Print how many keywords are contained in the description.

Sample Input:

1
5
she
he
say
shr
her
yasherhs

Sample Output:

3

题目链接

AC自动机模板裸题，注意此题用cin,cout,string不关同步会TLE。

AC代码:

#include <bits/stdc++.h>

const int maxn = 5e5 + 5;

struct AhoCorasickAutomaton {
    // 子节点记录数组
    int Son[maxn][26];
    int Val[maxn];
    // 失配指针Fail数组
    int Fail[maxn];
    // 节点数量
    int Tot;

    // Trie Tree初始化
    void TrieInit() {
        Tot = 0;
        memset(Son, 0, sizeof(Son));
        memset(Val, 0, sizeof(Val));
        memset(Fail, 0, sizeof(Fail));
    }

    // 计算字母下标
    int Pos(char X) {
        return X - 'a';
    }

    // 向Trie Tree中插入Str模式字符串
    void Insert(char Str[]) {
        int Cur = 0, Len = int(strlen(Str));
        for (int i = 0; i < Len; ++i) {
            int Index = Pos(Str[i]);
            if (!Son[Cur][Index]) {
                Son[Cur][Index] = ++Tot;
            }
            Cur = Son[Cur][Index];
        }
        Val[Cur]++;
    }

    // Bfs求得Trie Tree上失配指针
    void GetFail() {
        std::queue<int> Que;
        for (int i = 0; i < 26; ++i) {
            if (Son[0][i]) {
                Fail[Son[0][1]] = 0;
                Que.push(Son[0][i]);
            }
        }
        while (!Que.empty()) {
            int Cur = Que.front(); Que.pop();
            for (int i = 0; i < 26; ++i) {
                if (Son[Cur][i]) {
                    Fail[Son[Cur][i]] = Son[Fail[Cur]][i];
                    Que.push(Son[Cur][i]);
                }
                else {
                    Son[Cur][i] = Son[Fail[Cur]][i];
                }
            }
        }
    }

    // 询问Str中出现的模式串数量
    int Query(char Str[]) {
        int Len = int(strlen(Str));
        int Cur = 0, Ans = 0;
        for (int i = 0; i < Len; ++i) {
            Cur = Son[Cur][Pos(Str[i])];
            for (int j = Cur; j && ~Val[j]; j = Fail[j]) {
                Ans += Val[j];
                Val[j] = -1;
            }
        }
        return Ans;
    }
};

int N;
char Str[maxn << 1];
AhoCorasickAutomaton AC;

int main(int argc, char *argv[]) {
    AC.TrieInit();
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
        scanf("%s", Str);
        AC.Insert(Str);
    }
    AC.GetFail();
    scanf("%s", Str);
    printf("%d\n", AC.Query(Str));
    return 0;
}