以前做过的用的字典树,可是貌似现在再用超内存。。。。求解释。。。

问了LYN用的map函数做的,又去小小的学了map函数。。。。

http://wenku.baidu.com/view/0b08cece05087632311212ba.html感觉这个写的挺详细的

http://wenku.baidu.com/view/d140cfcca1c7aa00b52acb46.html这个的话挺有意思,帮助理解

题目描述

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.

输入

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.

输出

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

示例输入

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

示例输出

3

这个是代码

 #include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
#include<iostream>
#include<algorithm>
using namespace std ;
int main()
{
int m ;
while(scanf("%d",&m)&&(m != ))
{
int n ;
cin>>n ;
map<string,int>mp ;
string sh ;
for(int i = ; i <= m ; i++)
{
cin>>sh ;
transform(sh.begin(),sh.end(),sh.begin(),::tolower) ;
mp[sh] = ;
}
for(int j = ; j <= n ; j++)
{
cin>>sh ;
transform(sh.begin(),sh.end(),sh.begin(),::tolower) ;
if(mp[sh])
{
m--;
mp[sh] = ;
}
}
cout<<m<<endl ;
}
return ;
}

SDUT1500 Message Flood的更多相关文章

  1. Message Flood

    Message Flood Time Limit: 1500MS Memory limit: 65536K 题目描述 Well, how do you feel about mobile phone? ...

  2. Sicily 1194. Message Flood

    题目地址:1194. Message Flood 思路: 不区分大小写,先全部转化为小写,用stl提供的函数做会很方便. 具体代码如下: #include <iostream> #incl ...

  3. STL 之map解决 Message Flood(原字典树问题)

                                                                                      Message Flood Time ...

  4. 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 ...

  5. 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 ...

  6. oj1500(Message Flood)字典树

    大意:输入几个字符串,然后再输入几个字符串,看第一次输入的字符串有多少没有在后面的字符串中出现(后输入的字符串不一定出现在之前的字符串中) #include <stdio.h> #incl ...

  7. SDUT 1500-Message Flood(set)

    Message Flood Time Limit: 1500ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描写叙述 Well, how do you feel abo ...

  8. Eclipse 4.2 failed to start after TEE is installed

    ---------------  VM Arguments---------------  jvm_args: -Dosgi.requiredJavaVersion=1.6 -Dhelp.lucene ...

  9. 浅谈原始套接字 SOCK_RAW 的内幕及其应用(port scan, packet sniffer, syn flood, icmp flood)

    一.SOCK_RAW 内幕 首先在讲SOCK_RAW 之前,先来看创建socket 的函数: int socket(int domain, int type, int protocol); domai ...

随机推荐

  1. IPC with pipes, demo of 'popen'

    #include <stdio.h> #include <unistd.h> int main() { FILE* stream = popen ("sort&quo ...

  2. Linux下编译安装mysql-5.0.45.tar.gz

    安装环境:VMware9(桥接模式) + Linux bogon 2.6.32-642.3.1.el6.x86_64(查看linux版本信息:uname -a) 先给出MySQL For Linux ...

  3. linux传送文件至服务器

    scp安全文件拷贝(基于ssh的登陆)     1.你想把本地/home下的文件linux.tar.gz传送至远端服务器10.108.125.30,远端服务器的账号名为name,保存至服务器/home ...

  4. Jquery Slick幻灯片插件

    slick 是一个基于 jQuery 的幻灯片插件,具有以下特点: 支持响应式 浏览器支持 CSS3 时,则使用 CSS3 过度/动画 支持移动设备滑动 支持桌面浏览器鼠标拖动 支持循环 支持左右控制 ...

  5. 惠普M1005打印机无法自动进纸的问题

    惠普M1005打印机无法自动进纸的问题 问题起因 其实我也不太清楚是什么起因,前一天用的好好的惠普M1005打印机,在打印时没有直接打印,会弹出一个提示对话框,同时打印机显示屏上显示“load tra ...

  6. VC Dimension -衡量模型与样本的复杂度

    (1)定义VC Dimension: dichotomies数量的上限是成长函数,成长函数的上限是边界函数: 所以VC Bound可以改写成: 下面我们定义VC Dimension: 对于某个备选函数 ...

  7. 第25章 项目6:使用CGI进行远程编辑

    初次实现 25-1 simple_edit.cgi --简单的网页编辑器 #!D:\Program Files\python27\python.exeimport cgiform = cgi.Fiel ...

  8. 安装mysql 5.5.14 报错

    提示cmake nod foundyum install cmake 原因是曾经服务器安装过mysql数据库Installing MySQL system tables...101223 14:28: ...

  9. CPU原理

    cpu map 1.CPU的整体架构: 2.从CPU向内存 3.CPU和内存的关系图 4.CPU指令集 5.A+B 6.结果输入寄存器 7.寄存器中的临时存储,用来暂存B 8.将B传入寄存器 9.A会 ...

  10. spring中Bean的注入类型

    1.属性注入    即通过setXxx()方法注入Bean的属性值或依赖对象,由于属性注入方式具有可选择性和灵活性高的优点,因此属性注入是实际应用中最常采用的注入方式.    属性注入要求Bean提供 ...