南邮 OJ 1121 Message Flood

Message Flood

时间限制(普通/Java) : 2000 MS/ 6000 MS          运行内存限制 : 65536 KByte
总提交 : 411            测试通过 : 110 

比赛描述

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 messages 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 messages , 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 pre line . After that there are m lines of alphabetic string s ,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 .

样例输入

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

样例输出

3

题目来源

第九届中山大学程序设计竞赛预选题

#include<iostream>
#include<string>
#include<map>
using namespace std;

int main(){
	int n,m,count=0,i;
	string s;
	map<string,int>friends,senders;
	map<string,int>::iterator it;
	while(cin>>n && n){
		cin>>m;
		count = 0;
		friends.clear();
		senders.clear();
		while(n--){
			cin>>s;
			for(i=0;i<(int)s.size();++i){
				if(s[i]>='a' && s[i]<='z'){
					s[i] += 'A'-'a';
				}
			}
			++friends[s];
		}
		while(m--){
			cin>>s;
			for(i=0;i<(int)s.size();++i){
				if(s[i]>='a' && s[i]<='z'){
					s[i] += 'A'-'a';
				}
			}
			++senders[s];
		}
		for(it=friends.begin();it!=friends.end();++it){
			if(!senders[it->first]){
				++count;
			}
		}
	cout<<count<<endl;
	}
}