南邮- Message Flood

                        Message Flood

时间限制(普通/Java):2000MS/6000MS          运行内存限制:65536KByte
总提交:341            测试通过:91

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

 （容器）代码：

#include<iostream>#include<set>#include<string>using namespace std;int main(){    string str;    int m,n;    while(cin>>n&&n)    {        set<string> myset;        cin>>m;        for(int i=0;i<n;i++)        {            cin>>str;            for(int j=0;j<str.length();j++)                str[j]=tolower(str[j]);            myset.insert(str);        }        for(int i=0;i<m;i++)        {            cin>>str;            for(int j=0;j<str.length();j++)                str[j]=tolower(str[j]);            set<string>::iterator itor=myset.find(str);            if(itor!=myset.end())                myset.erase(str);        }        cout<<myset.size()<<endl;    }    return 0;}

代码：

#include <iostream>#include <string>#include <algorithm>//algorithm,用cmp和sort函数时要添加的头文件#include <memory.h>//用memset时要添加的头文件using namespace std;  bool cmp(string a, string b){    return a < b;   //这个是增序} string name[20001], tmp;   //都是全局变量了bool found[20001];int n, m;bool bi_search();  int main(){    int i,count;     while (cin >> n && n) //这句话特别有用，要会使用    {        cin >> m;        memset(found, false, 20001*sizeof(bool));// 没有了这句话会有wrong answer        for (i = 0; i < n; i++)        {            cin >> name[i];            for (int j = 0; j < name[i].length(); j++)                name[i][j] = tolower(name[i][j]); //将大写字母转化成小写字母        }        count = n;        sort(name, name+n, cmp);//注意使用方法        for (i = 0; i < m; i++)        {            cin >> tmp;            for (int j = 0; j < tmp.length(); j++)                tmp[j] = tolower(tmp[j]);             if (bi_search())                count--;        }        cout << count << endl;    }    return 0;} bool bi_search()//这个二分法写得很好，适当记下{    int start = 0, mid, end = n-1;    while (start <= end)    {        mid = (start+end)/2;          if (tmp == name[mid])        {            if (!found[mid])            {                found[mid] = true;                return true;            }            else                return false;        }        else if (tmp < name[mid])            end = mid-1;        else             start = mid+1;    }    return false;}