失配指针原理
使当前字符失配时跳转到另一段从root开始每一个字符都与当前已匹配字符段某一个后缀完全相同且长度最大的位置继续匹配，如同KMP算法一样，AC自动机在匹配时如果当前字符串匹配失败，那么利用失配指针进行跳转。由此可知如果跳转，跳转后的串的前缀必为跳转前的模式串的后缀，并且跳转的新位置的深度（匹配字符个数）一定小于跳之前的节点（跳转后匹配字符数不可能大于跳转前，否则无法保证跳转后的序列的前缀与跳转前的序列的后缀匹配）。所以可以利用BFS在Trie上进行失败指针求解。
简单来说
失败指针的作用就是将主串某一位之前的所有可以与模式串匹配的单词快速在Trie树中找出。
构建失败指针
构造失败指针的过程概括起来就一句话：设这个节点上的字母为C，沿着它父亲节点的失败指针走，直到走到一个节点，它的子结点中也有字母为C的节点。然后把当前节点的失败指针指向那个字母也为C的儿子。如果一直走到了root都没找到，那就把失败指针指向root。
具体操作起来
先把root加入队列(root的失败指针指向自己或者NULL)，这以后我们每处理一个点，就把它的所有儿子加入队列。
匹配
匹配过程分两种情况：
(1)当前字符匹配，表示从当前节点沿着树边有一条路径可以到达目标字符，此时只需沿该路径走向下一个节点继续匹配即可，目标字符串指针移向下个字符继续匹配；
(2)当前字符不匹配，则去当前节点失败指针所指向的字符继续匹配，匹配过程随着指针指向root结束。重复这2个过程中的任意一个，直到模式串走到结尾为止。
hdu2222 Keywords Search
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
1
5
she
he
say
shr
her
yasherhs
Sample Output
3
代码如下：
//http://blog.csdn.net/creatorx/article/details/71100840
#include <iostream>
#include <cstdio>
#include <cstring>
//ac 自动机
using namespace std;
const int maxn = 26;
char s[100005];
char keyword[55];
struct node
{
node * next[maxn];
node * fail;
int sum;//相应的个数
node()
{
for(int i = 0 ; i < maxn; i++)
next[i] = NULL;
fail = NULL;
sum = 0;
}
};
node * q[500005];
node * root;
void Build_Trie(char * s)
{
node * p = root;
int len = strlen(s);
for(int i = 0 ; i < len ; i++)
{
int tmp = s[i]-'a';
if(p->next[tmp] == NULL)
{
node *newnode = new node;
p->next[tmp] = newnode;
}
p = p->next[tmp];
}
p->sum++;
}
void build_fail_pointer()
{
int head = 0;
int tail = 0;
q[head++] = root;
node * p, *tmp;
while(head != tail)
{
tmp = q[tail++];
for(int i = 0 ; i < maxn; i++)
{
if(tmp->next[i] != NULL)
{
//所有第一层的点都要指向root
if(tmp == root)
{
tmp->next[i]->fail = root;
}
else
{
p = tmp->fail;
while(p != NULL)
{
if(p->next[i] != NULL)
{
tmp->next[i]->fail = p->next[i];
break;
}
p = p->fail;
}
if(p == NULL)
tmp->next[i]->fail = root;
}
q[head++] = tmp->next[i];
}
}
}
}
int query(node * root)
{
int i, v, cnt = 0;
node *p = root;
int len = strlen(s);
for(int i = 0 ; i < len; i++)
{
v = s[i]-'a';
while(p->next[v] == NULL && p!=root)
p = p->fail;
p = p->next[v];
if(p == NULL)
p = root;
node * tmp = p;
while(tmp != root)
{
if(tmp->sum >= 0)
{
cnt += tmp->sum;
tmp->sum = -1;
}
else
break;
tmp = tmp->fail;
}
}
return cnt;
}
int main()
{
int T, n;
scanf("%d",&T);
while(T--)
{
root = new node;
scanf("%d",&n);
for(int i = 0 ; i < n ; i++)
{
scanf("%s",keyword);
Build_Trie(keyword);
}
build_fail_pointer();
scanf("%s",s);
printf("%d
",query(root));
}
return 0;
}