HDU2222 AC自动机

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 59781    Accepted Submission(s): 19700

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

 

 

Author

Wiskey

 

 题意：

给出n个模板串和一个文本串，问文本串中出现了多少个模板串。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN=1000000;
int nod[MAXN+9][27],val[MAXN+9],f[MAXN+9],sz;
char s[MAXN+9];
bool vis[MAXN+9];
void init()
{
    sz=0;
    memset(nod[0],0,sizeof(nod));
    f[0]=val[0]=0;
    memset(vis,0,sizeof(vis));
}
void insert(char *s)
{
    int len=strlen(s),rt=0;
    for(int i=0;i<len;i++){
        int id=s[i]-'a';
        if(!nod[rt][id]){
            nod[rt][id]=++sz;
            memset(nod[sz],0,sizeof(nod[sz]));
            val[sz]=0;
        }
        rt=nod[rt][id];
    }
    val[rt]++;
}
void get_fail()
{
    queue<int>q;
    for(int i=0;i<26;i++){
        int u=nod[0][i];
        if(u) { q.push(u);f[u]=0; }
    }
    while(!q.empty()){
        int rt=q.front();q.pop();
        for(int i=0;i<26;i++){
            int u=nod[rt][i];
            if(!u){
                nod[rt][i]=nod[f[rt]][i];
                continue;
            }
            q.push(u);
            f[u]=nod[f[rt]][i];
        }
    }
}
int find(char *s)
{
    int len=strlen(s),rt=0,ans=0;
    for(int i=0;i<len;i++){
        vis[rt]=1;
        int id=s[i]-'a';
        int u=nod[rt][id];
        while(u&&!vis[u]){
            vis[u]=1;
            ans+=val[u];
            u=f[u];
        }
        rt=nod[rt][id];
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%s",s);
            insert(s);
        }
        get_fail();
        scanf("%s",s);
        printf("%d\n",find(s));
    }
    return 0;
}