HDU2222(AC自动机)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 56569    Accepted Submission(s): 18500

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

 

//2016.10.09
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int K = 26;
const int N = 500010;

struct node//AC自动机中的节点
{
    node *ch[K], *fail;
    int match;
    void Clear(){
        memset(this, 0, sizeof(node));
    }
}*que[N];

struct AC_automaton
{
    node nodes[N], *root, *superRoot, *cur;
    node* newNode(){
        cur->Clear();
        return cur++;
    }
    void Clear(){//初始化字典树
        cur = nodes;
        superRoot = newNode();
        root = newNode();
        root->fail = superRoot;
        for(int i = 0; i < K; i++)
              superRoot->ch[i] = root;
        superRoot->match = -1;
    }
    void Insert(char *s){//向字典树中插入一个字符串
        node *t = root;
        for(; *s; s++){
            int p = *s-'a';
            if(t->ch[p] == NULL)
                  t->ch[p] = newNode();
            t = t->ch[p];
        }
        t->match++;
    }
    void build(){//构建自动机
        int p = 0, q = 0;
        que[q++] = root;
        while(p != q){
            node *t = que[p++];
            for(int i = 0; i < K; i++){
                if(t->ch[i]){
                    t->ch[i]->fail = t->fail->ch[i];
                    que[q++] = t->ch[i];
                }else{
                    t->ch[i] = t->fail->ch[i];
                }
            }
        }
    }
    int run(char* s){//计算str串中模式串出现的次数
        int ans = 0;
        node *t = root;
        for(; *s; s++){
            int p = *s-'a';
            t = t->ch[p];
            for(node *u = t; u->match != -1; u = u->fail){
                ans += u->match;
                u->match = -1;
            }
        }
        return ans;
    }
};

char str[1000005];
AC_automaton ac;

int main()
{
    int n, T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        getchar();
        ac.Clear();
        for(int i = 0; i < n; i++){
            scanf("%s", str);
            ac.Insert(str);
        }
        ac.build();
        scanf("%s", str);
        printf("%d\n", ac.run(str));
    }

    return 0;
}