zoukankan      html  css  js  c++  java
  • HDU 2222 Keywords Search

    Keywords Search

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 25356    Accepted Submission(s): 8280


    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
    题目大意:输入一组字符串,最后一行输入一个字符串,问在最后一行的字符串中前面的字符串出现了几个。
    解题方法:AC自动机。
    #include <stdio.h>
    #include <iostream>
    #include <string.h>
    using namespace std;
    
    #define  N   500010
    char str[1000010];
    
    typedef struct node
    {
        int nCount;
        node *fail;
        node *next[26];
        node()
        {
            nCount = 0;
            fail = NULL;
            for (int i = 0; i < 26; i++)
            {
                next[i] = NULL;
            }
        }
    }TreeNode;
    
    TreeNode *Queue[N];
    
    void Insert(TreeNode *pHead, char Keyword[])
    {
        int nLen = strlen(Keyword);
        TreeNode *p = pHead;
        for (int i = 0; i < nLen; i++)
        {
            int index = Keyword[i] - 'a';
            if (p->next[index] == NULL)
            {
                p->next[index] = new TreeNode;
            }
            p = p->next[index];
        }
        p->nCount++;
    }
    
    void BuildAC(TreeNode *pRoot)
    {
        int head = 0, tail = 0;
        Queue[tail++] = pRoot;
        while(tail != head)
        {
            TreeNode *p = Queue[head++];
            for (int i = 0; i < 26; i++)
            {
                if (p->next[i] != NULL)
                {
                    if (p == pRoot)
                    {
                        p->next[i]->fail = pRoot;
                    }
                    else
                    {
                        TreeNode *temp = p->fail;
                        while(temp != NULL)
                        {
                            if (temp->next[i] != NULL)
                            {
                                p->next[i]->fail = temp->next[i];
                                break;
                            }
                            temp = temp->fail;
                        }
                        if (temp == NULL)
                        {
                            p->next[i]->fail = pRoot;
                        }
                    }
                    Queue[tail++] = p->next[i];
                }
            }
        }
    }
    
    int Query(TreeNode *pRoot)
    {
        int result = 0;
        int nLen = strlen(str);
        TreeNode *p = pRoot;
        for (int i = 0; i < nLen; i++)
        {
            int index = str[i] - 'a';
            while(p != pRoot && p->next[index] == NULL)
            {
                p = p->fail;
            }
            p = p->next[index];
            if (p == NULL)
            {
                p = pRoot;
            }
            TreeNode *temp = p;
            while(temp != pRoot && temp->nCount != -1)
            {
                result += temp->nCount;
                temp->nCount = -1;
                temp = temp->fail;
            }
        }
        return result;
    }
    
    void DeleteNode(TreeNode *pHead)
    {
        for (int i = 0; i < 26; i++)
        {
            if (pHead != NULL)
            {
                DeleteNode(pHead->next[i]);
            }
        }
        delete pHead;
    }
    
    int main()
    {
        char Keyword[55];
        int nCase;
        TreeNode *pHead = NULL;
        scanf("%d", &nCase);
        int n;
        while(nCase--)
        {
            scanf("%d", &n);
            pHead = new TreeNode;
            for (int i = 0; i < n; i++)
            {
                scanf("%s", Keyword);
                Insert(pHead, Keyword);
            }
            BuildAC(pHead);
            scanf("%s", str);
            printf("%d
    ", Query(pHead));
            DeleteNode(pHead);
        }
        return 0;
    }
  • 相关阅读:
    【microstation CE二次开发】不打开microstation,以COM方式启动Microstation
    【microstation CE二次开发】环境搭建
    Node安装与卸载命令汇总
    Maven进行clean时报错,解决方法
    Django 报ckeditor/skins/moono/skin.js 404
    Handler dispatch failed; nested exception is java.lang.AbstractMethodError: Method com/mchange/v2/c3p0/impl/NewProxyResultSet.isClosed()Z is abstract
    精准测试系列分享之一:JaCoCo 企业级应用的优缺点分析
    Java 中常见的细粒度锁实现
    JVM 的运行时数据区域分布
    Java 细粒度锁续篇
  • 原文地址:https://www.cnblogs.com/lzmfywz/p/3182150.html
Copyright © 2011-2022 走看看