AC自动机学习资料：
Keywords SearchTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39019Accepted Submission(s): 12578
Problem 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
15shehesayshrheryasherhs
Sample Output
3
本题求给出的单词，，在文本中出现了几个。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <queue>using namespace std;struct Trie{int next[500010][26], fail[500010], end[500010];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<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]);}}}int query(char buf[]){int len = strlen(buf);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");}}}ac;char buf[1000010];int main(){int T;int n;scanf("%d", &T);while (T–){scanf("%d", &n);ac.init();for (int i = 0; i < n; i++){scanf("%s", buf);ac.insert(buf);}ac.build();scanf("%s", buf);printf("%d\n", ac.query(buf));}return 0;}
即使爬到最高的山上，一次也只能脚踏实地地迈一步。
