hdu2222-AC自动机

原来这个玩意就是在trie上跑kmp啊，以前一直以为是什么难得东西。找失配指针，按要求统计个数，注意当匹配了一个串之后顺便顺着失配指针将前缀串一并统计。这是显然的，如果hello匹配成功，那么hell也算成功匹配。

在构造失配指针时，注意到当前节点的失配指针就是其父亲节点的失配指针（如果指针指向的节点含有当前节点相同的字符）

失配指针指向的是最长的一个后缀。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 71602    Accepted Submission(s): 24427

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

#include<bits/stdc++.h>
using namespace std;
const int MAX_N=500005;
const int MAX_SIG=26;
char T[1000006];
struct node{
      int next[MAX_SIG];
      int fail,cnt;
    }AC[MAX_N];
struct aho
{
    int size;
    int idx(char c){return c-'a';}
    void init()
    {
        for(int i=0;i<MAX_N;++i)
        {
            memset(AC[i].next,0,sizeof(AC[i].next));
            AC[i].fail=AC[i].cnt=0;
        }
        size=1;
    }

    void insert(char *S)
    {
        int n=strlen(S);
        int u=0;
        for(int i=0;i<n;++i){
            int c=idx(S[i]);
            if(!AC[u].next[c]) AC[u].next[c]=size++;
                u=AC[u].next[c];
        }
        AC[u].cnt++;
    }

    void build(){
        queue<int> q;
        AC[0].fail=-1;
        q.push(0);
        while(!q.empty()){
            int u=q.front();
            q.pop();
            for(int i=0;i<MAX_SIG;++i)
            {
                if(AC[u].next[i]){
                    if(!u) AC[AC[u].next[i]].fail=0;
                    else{
                        int v=AC[u].fail;
                        while(v!=-1){
                            if(AC[v].next[i]){
                                AC[AC[u].next[i]].fail=AC[v].next[i];
                                break;
                            }
                            v=AC[v].fail;
                        }
                        if(v==-1) AC[AC[u].next[i]].fail=0;
                    }
                    q.push(AC[u].next[i]);
                }
            }
        }
    }
    int Get(int u)
    {
        int res=0;
        while(u!=-1){
            res=res+AC[u].cnt;
            AC[u].cnt=0;
            u=AC[u].fail;
        }
        return res;
    }
    int match(char *S)
    {
        int n=strlen(S);
        int u=0,res=0;
        for(int i=0;i<n;++i)
        {
            int c=idx(S[i]);
            if(AC[u].next[c]) {u=AC[u].next[c];}
            else{
                int v=AC[u].fail;
                while(v!=-1&&AC[v].next[c]==0) v=AC[v].fail;
                if(v==-1) u=0;
                else u=AC[v].next[c];
            }
            if(AC[u].cnt)
                res=res+Get(u);
        }
        return res;
    }
};
int main()
{
    int t,i,n;
    char s[55];
    cin>>t;
    for(i=1;i<=t;++i)
    {
        aho a;
        a.init();
        cin>>n;
        for(int j=1;j<=n;++j){
            scanf("%s",s);
            a.insert(s);
        }
        scanf("%s",T);
        a.build();
        printf("%d
",a.match(T));
    }
    return 0;
}