Message Flood

Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

 
 

Description

Well, how do you feel about mobile phone? Your answer would probably be something like that "It's so convenient and benefits people a lot". However, If you ask Merlin this question on the New Year's Eve, he will definitely answer "What a trouble! I have to keep my fingers moving on the phone the whole night, because I have so many greeting message to send!" Yes, Merlin has such a long name list of his friends, and he would like to send a greeting message to each of them. What's worse, Merlin has another long name list of senders that have sent message to him, and he doesn't want to send another message to bother them Merlin is so polite that he always replies each message he receives immediately). So, before he begins to send message, he needs to figure to how many friends are left to be sent. Please write a program to help him. Here is something that you should note. First, Merlin's friend list is not ordered, and each name is alphabetic strings and case insensitive. These names are guaranteed to be not duplicated. Second, some senders may send more than one message to Merlin, therefore the sender list may be duplicated. Third, Merlin is known by so many people, that's why some message senders are even not included in his friend list.

Input

There are multiple test cases. In each case, at the first line there are two numbers n and m (1<=n,m<=20000), which is the number of friends and the number of messages he has received. And then there are n lines of alphabetic strings(the length of each will be less than 10), indicating the names of Merlin's friends, one per line. After that there are m lines of alphabetic strings, which are the names of message senders. The input is terminated by n=0.

Output

For each case, print one integer in one line which indicates the number of left friends he must send.

Sample Input

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

Sample Output

3

方法一:
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <algorithm>
#include <ctype.h> using namespace std; int main()
{
char str[30];
int len;
int n, m;
int i, j;
map<string,int>ma;
while(scanf("%d", &n)!=EOF)
{ if(n==0)
{
break;
}
scanf("%d%*c", &m );
ma.clear();
for(i=0; i<n; i++)
{
scanf("%s", str );
len = strlen(str );
for(j=0; j<len; j++)
{
str[j] = tolower( str[j] );
}
ma[str]++;
}
for(j=0; j<m; j++)
{
scanf("%s", str );
len = strlen(str);
for(i=0; i<len; i++)
{
str[i]=tolower(str[i]);
}
if(ma[str]>=1)
{n--;
ma[str]=0;
}
} printf("%d\n", n );
}
return 0;
}

方法二:

#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <algorithm>
#include <ctype.h>
using namespace std; int main()
{
char str[30];
int len;
int n, m;
int i, j;
// map<string,int>ma;
while(scanf("%d", &n)!=EOF)
{ if(n==0)
{
break;
}
scanf("%d%*c", &m );
map<string,int>ma;
for(i=0; i<n; i++)
{
scanf("%s", str );
len = strlen(str );
for(j=0; j<len; j++)
{
str[j] = tolower( str[j] );
}
ma[str]++;
}
for(j=0; j<m; j++)
{
scanf("%s", str );
len = strlen(str);
for(i=0; i<len; i++)
{
str[i]=tolower(str[i]);
}
ma.erase(str);
}
printf("%d\n", ma.size() );
}
return 0;
}

STL 之map解决 Message Flood(原字典树问题)的更多相关文章

  1. STL MAP及字典树在关键字统计中的性能分析

    转载请注明出处:http://blog.csdn.net/mxway/article/details/21321541 在搜索引擎在通常会对关键字出现的次数进行统计,这篇文章分析下使用C++ STL中 ...

  2. poj 2503 Babelfish(Map、Hash、字典树)

    题目链接:http://poj.org/bbs?problem_id=2503 思路分析: 题目数据数据量为10^5, 为查找问题,使用Hash或Map等查找树可以解决,也可以使用字典树查找. 代码( ...

  3. [POJ] #1002# 487-3279 : 桶排序/字典树(Trie树)/快速排序

    一. 题目 487-3279 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 274040   Accepted: 48891 ...

  4. 字典树应用 - poj1002

    字典树应用 - poj 1002 Description Businesses like to have memorable telephone numbers. One way to make a ...

  5. POJ 2418 字典树

    题目链接:http://poj.org/problem?id=2418 题意:给定一堆树的名字,现在问你每一棵树[无重复]的出现的百分比,并按树名的字典序输出 思路:最简单的就是用map来写,关于字典 ...

  6. Go语言字典树定义及实现

    // trie 字典树实现 package Algorithm // 字典树节点 type TrieNode struct { children map[interface{}]*TrieNode i ...

  7. sdut Message Flood(c++ map)

    用字典树没过,学习了一下map; 参考博客:http://blog.csdn.net/zhengnanlee/article/details/8962432 AC代码 #include<iost ...

  8. Message Flood(map)

    http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=203#problem/D 以前用字典树做过 #include <strin ...

  9. stl应用(map)或字典树(有点东西)

    M - Violet Snow Gym - 101350M Every year, an elephant qualifies to the Arab Collegiate Programming C ...

随机推荐

  1. jquery的push()

    JavaScript push() 方法 JavaScript Array 对象 定义和用法 push() 方法可向数组的末尾添加一个或多个元素,并返回新的长度. 语法 arrayObject.pus ...

  2. [译]GLUT教程 - 键盘

    Lighthouse3d.com >> GLUT Tutorial >> Input >> Keyboard GLUT可以让应用程序自动监测键盘输入,包括普通按键和 ...

  3. css3 jQuery实现3d搜索框+为空推断

    <!DOCTYPE html> <html> <head> <title>css3实现3d搜索框</title> <style> ...

  4. 【JMeter4.0学习(二)】之搭建openLDAP在windows8.1上的安装配置以及JMeter对LDAP服务器的性能测试脚本开发

    目录: 概述 安装测试环境 安装过程 配置启动 配置搭建OpenLDAP 给数据库添加数据 测试查询刚刚插入的数据 客户端介绍 JMeter建立一个扩展LDAP服务器的性能测试脚本开发 附:LDAP学 ...

  5. laravel学习之路1:认证相关

    Laravel中Auth::guard()表示什么意思? Auth::check() 是判断用户是否登录的方法,如果使用的默认用户系统,那这样使用没问题. 但是使用两组用户的话,如何使用各组用户的功能 ...

  6. CentOS VSCode调试go语言出现:exec: "gcc": executable file not found in PATH

    CentOS VSCode调试go语言出现:exec: "gcc": executable file not found in PATH 解决方案: 执行如下命令安装GCC,然后重 ...

  7. mysql-5.1.73多实例安装启动

    一.源码包下载:http://download.softagency.net/MySQL/Downloads/MySQL-5.1/ 二.编译安装 groupadd mysql useradd -r - ...

  8. iOS 线程管理的学习记录

    本文转载至 http://www.2cto.com/kf/201312/265451.html 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 2 ...

  9. [转]Html position(static、relative、absolute、fixed)

    转自:http://blog.csdn.net/topviewers/article/details/21644305 讲解不错,转载备忘. position的四个属性值: 1.relative2.a ...

  10. Android笔记之使用Glide加载网络图片、下载图片

    Glide简介 不想说太多,真的很方便:P)可以节省我不少时间 GitHub地址:https://github.com/bumptech/glide 加载网络图片到ImageView Glide.wi ...