Message Flood
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;
}