本文共 4596 字，大约阅读时间需要 15 分钟。

AC自动机

用于解决多模式串匹配

#include 
   
    using namespace std;const int maxn = 1e5 + 7;struct Trie {
       int next[maxn][26], fail[maxn], end[maxn];    int root, L;    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 insert(char *buf) {
           int len = strlen(buf);        int now = root;        for (int i = 0; i < len; i++) {
               if (next[now][buf[i] - 'a'] == -1)                next[now][buf[i] - 'a'] = newnode();            now = next[now][buf[i] - 'a'];        }        end[now]++;    }    void build() {
           queue
    
      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]);                }        }    }    int query(string buf) {
           int len = buf.length();        int now = root;        int res = 0;        for (int i = 0; i < len; i++) {
               now = next[now][buf[i] - 'a'];            int temp = now;            while (temp != root) {
                   res += end[temp];                end[temp] = 0;                temp = fail[temp];            }        }        return res;    }    void debug() {
           for (int i = 0; i < L; i++) {
               printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], end[i]);            for (int j = 0; j < 26; j++)                printf("%2d", next[i][j]);            printf("]\n");        }    }};char str[maxn];Trie ac;string pat;
    
   

例题：HDU 2222

第一行输入测试数据的组数，然后输入一个整数n，接下来的n行每行输入一个单词，最后输入一个字符串，问在这个字符串中有多少个单词出现过。

求目标串中出现了几个模式串

#include 
   
    using namespace std;const int maxn = 1e5 + 7;struct Trie {
       int next[maxn][26], fail[maxn], end[maxn];    int root, L;    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 insert(char *buf) {
           int len = strlen(buf);        int now = root;        for (int i = 0; i < len; i++) {
               if (next[now][buf[i] - 'a'] == -1)                next[now][buf[i] - 'a'] = newnode();            now = next[now][buf[i] - 'a'];        }        end[now]++;    }    void build() {
           queue
    
      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]);                }        }    }    int query(string buf) {
           int len = buf.length();        int now = root;        int res = 0;        for (int i = 0; i < len; i++) {
               now = next[now][buf[i] - 'a'];            int temp = now;            while (temp != root) {
                   res += end[temp];                end[temp] = 0;                temp = fail[temp];            }        }        return res;    }    void debug() {
           for (int i = 0; i < L; i++) {
               printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], end[i]);            for (int j = 0; j < 26; j++)                printf("%2d", next[i][j]);            printf("]\n");        }    }};char str[maxn];Trie ac;string pat;int main() {
       int T;    scanf("%d", &T);    while (T--) {
           int n;        ac.init();        scanf("%d", &n);        pat.clear();        for (int i = 0; i < n; i++) {
               scanf("%s", str);            ac.insert(str);            if (strlen(str) > pat.size()) {
                   pat = str;            }        }        cin >> pat;        ac.build();        int ans = ac.query(pat);        printf("%d\n", ans);    }    return 0;}
    
   

转载地址：https://blog.csdn.net/weixin_43820352/article/details/108523156 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！