题目：给你一些单词构成的词典，统计出现过的单词个数。

分析：字符串，AC自动机。比较裸的AC自动机详细可参照本空间的：AC自动机总结。

说明：注意每个单词只统计一次。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* AC_DFA define */
#define nodesize 250010        //节点个数
#define dictsize 26            //符集大小 

typedef struct node1
{
    int    flag;            //值域 
    node1* fail;
    node1* next[dictsize];
}tnode;
tnode  dict[nodesize+1];
tnode* Q[nodesize+1];
int    ID[256]; 

class AC_DFA
{
    private:
        int    size;
        tnode* root;
    public:
        AC_DFA() {
            makeID();
            memset( dict, 0, sizeof( dict ) );
            root=NULL; size=0; root=newnode();
        }
        void makeID() {
            for ( int i = 0 ; i < 26 ; ++ i )
                ID['a'+i] = i;
        }
        void init() {
            memset( dict, 0, sizeof( dict ) );
            root=NULL; size=0; root=newnode();
        }
        tnode* newnode() {
            dict[size].fail = root;
            return &dict[size ++];
        }
        void insert( char* word, int l ) {
            tnode* now = root;
            for ( int i = 0 ; i < l ; ++ i ) {
                if ( !now->next[ID[word[i]]] )
                    now->next[ID[word[i]]] = newnode();
                now = now->next[ID[word[i]]];
            }now->flag ++;
        }
        void setfail() {
            Q[0] = root; root->fail = NULL;
            for ( int move = 0,save = 1 ; move < save ; ++ move ) {
                tnode* now = Q[move];
                for ( int i = 0 ; i < dictsize ; ++ i )
                    if ( now->next[i] ) {
                        tnode* p = now->fail;
                        while ( p && !p->next[i] ) p = p->fail;
                        now->next[i]->fail = p?p->next[i]:root;
                        Q[save ++] = now->next[i];
                    }else now->next[i] = now==root?root:now->fail->next[i];//构建 Trie图 
            }
        }
        int query( char* line, int L ) {//统计字串出现个数，可重复及交叉
            int sum = 0;
            tnode *temp,*now = root;
            for ( int i = 0 ; i < L ; ++ i ) {
                now = now->next[ID[line[i]]];
                temp = now;
                while (temp && temp->flag) {
                	sum += temp->flag;
					temp->flag = 0;
					temp = temp->fail;
				}
            }
            return sum; 
        }
};
/* AC_DFA  end */

char Word[51];
char Line[1000001];

int main()
{
    int T,N;
    while ( ~scanf("%d",&T) ) 
    for ( int t = 1 ; t <= T ; ++ t ) {
        AC_DFA ac;
        scanf("%d",&N);
        for ( int i = 1 ; i <= N ; ++ i ) {
            scanf("%s",Word);
            ac.insert( Word, strlen( Word ) );
        }
        ac.setfail();
        scanf("%s",Line);
        printf("%d\n",ac.query( Line, strlen(Line) ));
    }
    return 0;
}

hdu 2222 - Keywords Search