题意：给定n个模式串和一个文本串，判断文本中是否存在模式串。

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思路：套模板即可。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1e6 + 5;
struct Aho{
    struct State{
        int next[26];
        int fail, cnt;
    }state[maxn];
    queue<int>que;
    int size;

    void init() {
        size = 1;
        while(!que.empty()) que.pop();
        for(int i = 0; i < maxn; ++i) {
            memset(state[i].next, 0, sizeof(state[i].next));
            state[i].cnt = state[i].fail = 0;
        }
    }

    void insert(char *S) {
        int n = strlen(S);
        int now = 0;
        for(int i = 0; i < n; ++i) {
            int num = S[i]-'a';
            if(state[now].next[num]) now = state[now].next[num];
            else {
                state[now].next[num] = size++;
                now = state[now].next[num];
            }
        }
        state[now].cnt++;
    }

    void getFail() {
        state[0].fail = -1;
        que.push(0);
        while(!que.empty()) {
            int u = que.front(); que.pop();
            for(int i = 0; i < 26; ++i) {
                if(state[u].next[i]) {
                    if(u == 0) state[state[u].next[i]].fail = 0;
                    else {
                        int v = state[u].fail;
                        while(v != -1 && !state[v].next[i]) v = state[v].fail;
                        if(v == -1) state[state[u].next[i]].fail = -1;
                        else state[state[u].next[i]].fail = state[v].next[i];

                    }
                    que.push(state[u].next[i]);
                }
            }
        }
    }

    int Get(int u) {
        int res = 0;
        while(u) {
            res += state[u].cnt;
            state[u].cnt = 0;
            u = state[u].fail;
        }
        return res;
    }

    bool match(char *S) {
        int n = strlen(S);
        int now = 0, res = 0;
        for(int i = 0; i < n; ++i) {
            int num = S[i]-'a';
            if(state[now].next[num]) now = state[now].next[num];
            else {
                int u = state[now].fail;
                while(u != -1 && !state[u].next[num]) u = state[u].fail;
                if(u == -1) now = 0;
                else now = state[u].next[num];
            }
            if(state[now].cnt) res += Get(now);
            if(res) return true;
        }
        return false;
    }
}aho;

char S[maxn];
int n;
int main() {
    while(scanf("%d", &n) == 1) {
        aho.init();
        for(int i = 0; i < n; ++i) {
            scanf("%s", S);
            aho.insert(S);
        }
        aho.getFail();
        scanf("%s", S);
        if(aho.match(S)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

如有不当之处欢迎指出！