zoukankan      html  css  js  c++  java
  • hdu 3695 AC自动机模板题

    Keywords Search

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 23502    Accepted Submission(s): 7763


    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
     

    ---------------------

    模板题,不能更水。

    ---------------------

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    #include <queue>
    #include <cstring>
    
    using namespace std;
    
    //子树节点是在插入时new的,
    //寻找失配指针中使用的队列是直接调用STL的
    const int kind = 26;
    struct node
    {
        node *fail;
        node *next[kind];
        int count;//记录当前前缀是完整单词出现的个数
        node()
        {
            fail = NULL;
            count = 0;
            memset(next,NULL,sizeof(next));
        }
    };
    
    void insert(char *str,node *root)
    {
        node *p=root;
        int i=0,index;
        while(str[i])
        {
            index = str[i]-'a';
            if(p->next[index]==NULL) p->next[index]=new node();
            p=p->next[index];
            i++;
        }
        p->count++;
    
    }
    
    //寻找失败指针
    void build_ac_automation(node *root)
    {
        int i;
        queue<node *>Q;
        root->fail = NULL;
        Q.push(root);
        while(!Q.empty())
        {
            node *temp = Q.front();//q[head++];//取队首元素
            Q.pop();
            node *p = NULL;
            for(i=0; i<kind; i++)
            {
                if(temp->next[i]!=NULL)//寻找当前子树的失败指针
                {
                    p = temp->fail;
                    while(p!=NULL)
                    {
                        if(p->next[i]!=NULL)//找到失败指针
                        {
                            temp->next[i]->fail = p->next[i];
                            break;
                        }
                        p = p->fail;
                    }
    
                    if(p==NULL)//无法获取,当前子树的失败指针为根
                        temp->next[i]->fail = root;
    
                    Q.push(temp->next[i]);
                }
            }
        }
    }
    
    
    //询问str中包含n个关键字中多少种即匹配
    int query(char *str,node *root)
    {
        int i = 0,cnt = 0,index,len;
        len = strlen(str);
        node *p = root;
        while(str[i])
        {
            index = str[i]-'a';
            while(p->next[index]==NULL&&p!=root)//失配
                p=p->fail;
            p=p->next[index];
            if(p==NULL)//失配指针为根
                p = root;
    
            node *temp = p;
            while(temp!=root&&temp->count!=-1)//寻找到当前位置为止是否出现病毒关键字
            {
                //if(temp->count!=0)
                cnt+=temp->count;
                temp->count=-1;
                temp=temp->fail;
            }
            i++;
        }
        return cnt;
    }
    
    char str[1111111];
    char words[1111111];
    int T,n;
    
    node* root;
    
    int main()
    {
        scanf("%d",&T);
        while (T--)
        {
            scanf("%d",&n);
            root=new node();
            while (n--)
            {
                scanf("%s",words);
                insert(words,root);
            }
            build_ac_automation(root);
            scanf("%s",str);
            int ans=query(str,root);
            printf("%d\n",ans);
        }
        return 0;
    }
    





  • 相关阅读:
    js汉字转换为阿拉伯数字支持十到十九
    JS中判断是中文数字的函数
    一个JS正则的字符串替换函数
    thinkphp3.2.3使用formdata的多文件上传
    计算列表中的名字出现的订单中的订单总额
    配置ssl使用了不受支持的协议。 ERR_SSL_VERSION_OR_CIPHER_MISMATCH
    IIS8.5中的强制https直接修改web.config文件和顶级域名跳转www和过滤子目录不强制跳转
    关于wordpress4.8中的Twenty Seventeen主题的主题选项增加章节的实现
    excel中统计COUNTIFS的值为0
    thinkphp3.2.3集成phpexcel1.8导出设置单元格合并
  • 原文地址:https://www.cnblogs.com/cyendra/p/3038378.html
Copyright © 2011-2022 走看看