HDOJ 2222 Keywords Search

Keywords Search

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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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 

 

一直觉得网上没有顺眼的AC自动机，所以参照白书写了个AC自动机的模板。。。。

AC自动机主要就3个部分：1 建树：和普通字典树一样

                                               2 建失败指针：

                                                ///根结点的失败指针指向自己
                                               ///每个结点的失败指针是从父节点的失败指针指向的结点的子结点中寻找与这个结点相同的结点
                                               ///如果找不到，就向上回溯直到根节点

                                               同时 顺手把不存在的边都给补上，顺手建last数组

                                              3 模式匹配：因为失配函数的时候顺手把不存在的边都给补上了，直接顺着失配边走下去就好了

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn=550000;struct AC_auto{    int chd[maxn][26],v[maxn],f[maxn],last[maxn],sz,ans;    void init()    {        sz=1;ans=0;        memset(v,0,sizeof(v));        memset(f,0,sizeof(f));        memset(chd[0],0,sizeof(chd[0]));    }    void insert(char* p)    {        int cur=0;        for(;*p;p++)        {            if(!chd[cur][*p-'a'])            {                memset(chd[sz],0,sizeof(chd[sz]));                chd[cur][*p-'a']=sz++;            }            cur=chd[cur][*p-'a'];        }        v[cur]++;    }    bool query(char* p)    {        int cur=0;        for(;*p;p++)        {            if(!chd[cur][*p-'a']) break;            cur=chd[cur][*p-'a'];        }        return v[cur]&&(!(*p));    }    int getFail()    {        queue<int> q;        f[0]=0;        for(int c=0;c<26;c++)        {            int u=chd[0][c];            if(u)            {                f[u]=0; q.push(u); last[u]=0;            }        }        while(!q.empty())        {            int r=q.front(); q.pop();            for(int c=0;c<26;c++)            {                int u=chd[r][c];                if(!u){ chd[r][c]=chd[f[r]][c];continue;}///....                q.push(u);                int vv=f[r];                while(vv&&!chd[vv][c]) vv=f[vv];                f[u]=chd[vv][c];                last[u]=v[f[u]] ? f[u] : last[f[u]];            }        }    }    void solve(int j)    {        if(!j) return;        if(v[j])        {            ans+=v[j];            v[j]=0;        }        solve(last[j]);    }    void find(char* T)    {        int n=strlen(T),j=0;        getFail();        for(int i=0;i<n;i++)        {            //while(j&&!chd[j][*T-'a']) j=f[j];            j=chd[j][T[i]-'a'];            if(v[j]) solve(j);            else if(last[j]) solve(last[j]);        }    }}ac;int main(){    int t,n;    char dic[100],str[1100000];    scanf("%d",&t);    while(t--)    {        ac.init();        scanf("%d",&n);        while(n--)        {            scanf("%s",dic);            ac.insert(dic);        }        scanf("%s",str);        ac.find(str);        printf("%d\n",ac.ans);    }    return 0;}