hdu 2222 Keywords Search（ac自动机模板题，但是有陷阱）

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27477    Accepted Submission(s): 9024

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

 

Recommend

lcy

 

题意：

多模板匹配问题。给你n个模板串和一个文本串。问你有几个模板串在文本串中出现过。

思路：

这题有个陷阱。就是模板串会给你重复的！巨坑！而且要把它们当不同的串处理！也就是输入了几次就算几个串。

所以用val记录单词出现次数就好。匹配一次后就把val归0.用ans统计。

详细见代码：

#include <iostream>
#include<queue>
#include<map>
#include<string.h>
#include<stdio.h>
using namespace std;
const int md=500110;
const int ssz=26;
const int maxn=1000010;
char txt[maxn];
char words[60];
int ch[md][ssz],val[md],f[md],last[md];
int sz,ans;
void init()
{
    sz=1;
    val[0]=0;
    memset(ch[0],0,sizeof ch[0]);
}
int idx(char c)
{
    return c-'a';
}
void inser(char *s,int v)
{
    int u=0,n=strlen(s);
    for(int i=0; i<n; i++)
    {
        int c=idx(s[i]);
        if(!ch[u][c])
        {
            memset(ch[sz],0,sizeof ch[sz]);
            val[sz]=0;
            ch[u][c]=sz++;
        }
        u=ch[u][c];
    }
    val[u]++;//同一单词可以多次计算。。。坑啊。。。
}
void solve(int j)
{
    if(j)
    {
        if(val[j])
        {
            ans+=val[j];
            val[j]=0;
        }
        solve(last[j]);
    }
}
void getf()
{
    queue<int> q;
    int u,v,c,r;
    f[0]=0;
    for(c=0;c<ssz;c++)
    {
        u=ch[0][c];
        if(u)
        {
            f[u]=0;
            q.push(u);
            last[u]=0;
        }
    }
    while(!q.empty())//bfs获取失配数组
    {
        r=q.front();
        q.pop();
        for(c=0;c<ssz;c++)
        {
            u=ch[r][c];
            if(!u)
            {
                ch[r][c]=ch[f[r]][c];//直接增加失配边
                continue;
            }
            q.push(u);
            v=f[r];
            while(v&&!ch[v][c])
                v=f[v];
            f[u]=ch[v][c];
            last[u]=val[f[u]]?f[u]:last[f[u]];
        }
    }
}
void findm(char *T)
{
    int n=strlen(T),c,i,j=0;
    getf();//获得失配函数
    for(i=0;i<n;i++)
    {
        c=idx(T[i]);
        //while(j&&!ch[j][c])
        //    j=f[j];
        j=ch[j][c];//沿着边走就行
        if(val[j])
            solve(j);
        else if(last[j])
            solve(last[j]);
    }
}
int main()
{
    int i,t,n;

    scanf("%d",&t);
    while(t--)
    {
        init();
        ans=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%s",words);
            inser(words,i);
        }
        scanf("%s",txt);
        findm(txt);
        printf("%d\n",ans);
    }
    return 0;
}
/*
1
h 
hhhhh 
2 
a 
a 
a 
*/
/*
1
2
*/