zoukankan      html  css  js  c++  java
  • HDU-2222

    • Keywords Search

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 39064    Accepted Submission(s): 12596


    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
     
    Recommend

    lcy

    题意:看现在的主字符串,有多少个零碎的字符串构成,
    做法:AC自动机模板
    代码如下:
    #include<iostream>
    #include<string.h>
    #include<algorithm>
    #include<stdio.h>
    #include<cmath>
    #include<queue>
    #define maxn 500010
    using namespace std;
    struct Tire
    {
        int next[maxn][26],fail[maxn],end[maxn];
        int root,L;
        int newnode()
        {
            for(int i=0; i<26; i++)
            {
                next[L][i] = -1;
            }
            end[L++] = 0;
            return L-1;
        }
        void init()
        {
                        L = 0;
                        root = newnode();
        }
        void insert(char buf[])
        {
                        int len = strlen(buf);
                        int now = root;
                        for(int i=0;i<len;i++)
                        {
                                  if(next[now][buf[i] - 'a'] == -1)
                                  {
                                            next[now][buf[i] -'a'] = newnode();
                                  }
                                  now = next[now][buf[i] -'a'];
                        }
                        end[now]++;
        }
        void build()
        {
                        queue<int>Q;
                        fail[root] = root;
                        for(int i=0;i<26;i++)
                        {
                                  if(next[root][i] == -1)
                                  next[root][i] = root;
                                  else
                                  {
                                            fail[next[root][i]] = root;
                                            Q.push(next[root][i]);
                                  }
                        }
                        while(!Q.empty())
                        {
                                  int now = Q.front();
                                  Q.pop();
                                  for(int i=0;i<26;i++)
                                  {
                                           if(next[now][i] == -1)
                                           next[now][i] = next[fail[now]][i];
                                           else
                                           {
                                            fail[next[now][i]] = next[fail[now]][i];
                                            Q.push(next[now][i]);
                                           }
                                  }
                        }
        }
        int query(char buf[])
        {
                        int len = strlen(buf);
                        int now = root;
                        int res = 0;
                        for(int i=0;i<len;i++)
                        {
                                  now = next[now][buf[i] -'a'];
                                  int tmp = now;
                                  while(tmp != root)
                                  {
                                            res += end[tmp];
                                            end[tmp] = 0;
                                            tmp = fail[tmp];
                                  }
                        }
                        return res;
        }
    };
    char buf[1000010];
    Tire ac;
    int main()
    {
    #ifndef ONLINE_JUDGE
              freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
              int T;
              scanf("%d",&T);
              while(T--)
              {
                        int n;
                        scanf("%d",&n);
                        ac.init();
                        for(int i=0;i<n;i++)
                        {
                                  scanf("%s",buf);
                                  ac.insert(buf);
                        }
                        ac.build();
                        scanf("%s",buf);
                        printf("%d
    ",ac.query(buf));
              }
              return 0;
    }
  • 相关阅读:
    eclipse中svn的各种状态图标详解
    Invalid configuation file. File "**********" was created by a VMware product with more feature than this version of VMware Workstation and cannot be
    linux下tomcat无法访问问题(换一种说法:无法访问8080端口)
    安装MySQL start Service(无法启动服务)
    eclipse下SVN subclipse插件
    tomcat启动窗口中的时间与系统时间不一致
    关于如果从SQLSERVER中获取 数据库信息 或者 表信息
    有关google的appengine部署服务器的简单教程
    部署到Google App Engine时中途退出后引起的问题
    重温WCF之数据契约中使用枚举(转载)(十一)
  • 原文地址:https://www.cnblogs.com/chenyang920/p/4355369.html
Copyright © 2011-2022 走看看