HDU - 2222 Keywords Search

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.
InputFirst 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.
OutputPrint how many keywords are contained in the description.Sample Input

1
5
she
he
say
shr
her
yasherhs

Sample Output

3

AC自动机裸题，下面给出用指针实现和用数组实现两种方法。指针比较容易理解，但数组不会出现各种奇奇怪怪的玄学问题。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <algorithm>

using namespace std;
char str[1000000+100];

struct node
{
    int num;
    struct node *next[26];
    struct node *fail;
    void init()
    {
        int i;
        for(i=0;i<26;i++)
            next[i]=NULL;
        num=0;
        fail=NULL;
    }
}*root;


node* rt;
int id,len;

void build()
{
    rt=root;
    len=strlen(str);
    for(int i=0;i<len;i++)
    {
        id=str[i]-'a';
        if(rt->next[id]==NULL)
        {
            rt->next[id]=new node();
            rt->next[id]->init();
        }    
        rt=rt->next[id];
    }
    rt->num++;
}

void get_fail()
{
    rt=root;
    node *son,*temp;
    queue<node *> q;
    q.push(rt);
    while(!q.empty())
    {
        temp=q.front();
        q.pop();
        for(int i=0;i<26;i++)
        {
            son=temp->next[i];
            if(son!=NULL)
            {
                if(temp==root)
                    son->fail=root;
                    else
                    {
                        rt=temp->fail;
                        while(rt)
                        {
                            if(rt->next[i])
                            {
                                son->fail=rt->next[i];
                                break;
                            }
                            rt=rt->fail;
                        }
                        if(!rt)
                            son->fail=root;
                    }
                q.push(son);
            }
        }
    }
}

int querry()
{
    rt=root;
    len=strlen(str);
    node *temp;
    int ans=0;
    for(int i=0;i<len;i++)
    {
        id=str[i]-'a';
        while(!rt->next[id]&&rt!=root)
            rt=rt->fail;
        rt=rt->next[id];
          if(!rt)
              rt=root;
          temp=rt;
          while(temp!=root)
          {
              if(temp->num>=0)
            {
                ans+=temp->num;
                temp->num=-1;    
            }    
            else
                break;
            temp=temp->fail;
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        root=new node;
        root->init();
        scanf("%d",&n);
        getchar();
        for(int i=0;i<n;i++)
        {
            gets(str);
            build();            
        }
        get_fail();
        gets(str);
        printf("%d\n",querry());    
    }
    return 0;
}

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;

const int N=10100,L=1000100;
char s[L];
int num,n;
struct node{
    int son[30];
    int fail,cnt;
}a[N*60];
queue<int> q;

void clear(int x)
{
    a[x].cnt=0;
    a[x].fail=0;
    memset(a[x].son,0,sizeof(a[x].son));
}

void trie(char *c)
{
    int l=strlen(c);
    int x=0;
    for(int i=0;i<l;i++)
    {
        int t=c[i]-'a'+1;
        if(!a[x].son[t])
        {
            num++;
            clear(num);
            a[x].son[t]=num;
        }
        x=a[x].son[t];
    }
    a[x].cnt++;
}

void buildAC()
{
    while(!q.empty()) q.pop();
    for(int i=1;i<=26;i++)
        if(a[0].son[i]) q.push(a[0].son[i]);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        int fail=a[x].fail;
        for(int i=1;i<=26;i++) 
        {
            int y=a[x].son[i];
            if(y)
            {
                a[y].fail=a[fail].son[i];
                q.push(y);
            }
            else a[x].son[i]=a[fail].son[i];
        }
    }
}

int find(char *c)
{
    int l=strlen(c);
    int x=0,ans=0;
    for(int i=0;i<l;i++)
    {
        int t=c[i]-'a'+1;
        while(x && !a[x].son[t]) x=a[x].fail;        
        x=a[x].son[t];
        int p=x;
        while(p && a[p].cnt!=-1)
        {
            ans+=a[p].cnt;
            a[p].cnt=-1;
            p=a[p].fail;
        }
    }
    return ans;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        num=0;
        clear(0);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            trie(s);
        }
        buildAC();
        scanf("%s",s);
        printf("%d\n",find(s));
    }
    return 0;
}