HDU2222 Keywords Search（AC自动机）

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 46676    Accepted Submission(s): 14858

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

 

【思路】

       AC自动机模板的直接应用。

       需要注意的有：

　　　　1)      要用val纪录该节点对应得单词数目，因为可能会有重复的串。

　　　　2)      Val累计入cnt后清0，print可能会重复遍历一个相同的单词。

【代码】

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
#define FOR(a,b,c) for(int a=(b);a<=(c);a++)
using namespace std;

const int maxn = 10000+10;
const int maxl = 50+5;
const int maxnode = maxl*maxn;
const int sigma = 26;

struct ACautomaton {
    int ch[maxnode][sigma];
    int f[maxnode],val[maxnode],last[maxnode];
    int sz,cnt;
    
    void clear() {
        sz=1; cnt=0;
        memset(ch[0],0,sizeof(ch[0]));
        val[0]=f[0]=0;
    }
    int idx(char c) {  return c-'a';  }
    void insert(char* s) {
        int n=strlen(s),u=0;
        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]++;
    }
    void print(int j) {
        if(j) {
            cnt+=val[j];
            val[j]=0;                //一定要置为0防止重复累计 
            print(last[j]);
        }
    }
    void find(char* s) {
        int n=strlen(s),j=0;
        for(int i=0;i<n;i++) {
            int c=idx(s[i]);
            while(j && !ch[j][c]) j=f[j];
            j=ch[j][c];
            if(val[j]) print(j);
            else if(last[j]) print(last[j]);
        }
    }
    void get_Fail() {
        queue<int> q;
        f[0]=0;
        for(int c=0;c<sigma;c++) {
            int u=ch[0][c];
            if(u)  {  f[u]=last[u]=0; q.push(u);  }
        }
        while(!q.empty()) {
            int r=q.front(); q.pop();
            for(int c=0;c<sigma;c++) {
                int u=ch[r][c];
                if(!u) continue;
                q.push(u);
                int 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]];
            }
        }
    }
}ac;

int n;

int main() {
    //freopen("in.in","r",stdin);
    //freopen("out_me.out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--) {
        char s[maxl];
        scanf("%d",&n);
        ac.clear();
        FOR(i,1,n) {
            scanf("%s",s);
            ac.insert(s);
        }
        ac.get_Fail();
        char T[1000010];
        scanf("%s",T);
        ac.find(T);
        printf("%d
",ac.cnt);
    }
    return 0;
}