sicily-1194. Message Flood

1194. Message Flood

限制条件
时间限制: 1 秒, 内存限制: 32 兆

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 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.


输入格式

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.



输出格式

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
ZSUACM Team Member

 

这道题由多组数据，每组数据第一行输入为n,m（0<=n,m <=20000),其中n为朋友的个数，m为收到短信的条数，接下来有n+m个名字，求剩下的朋友的个数。

思路比较简单，用二份查找和 sort排序就行。下面是代码

//1194. Message Flood//2016.12.07#include <iostream>#include <string>#include <algorithm>#include <cstring>#include <cstdlib>using namespace std;string name[20005];  // 存储朋友名字 bool send[20005];   // 判断是否有短信 string tmp;        // 发了短信的名字 int n,m;bool binary()      // 二分查找 {int begin,mid,end;begin = 0;end = n-1;while(begin<=end){mid = (begin+end)/2;if(tmp == name[mid]){if(!send[mid]){send[mid]=true;return true;}elsereturn false;}else if(tmp < name[mid])end = mid-1;elsebegin = mid+1;}return false;}bool cmp(string a,string b){return a < b;}int main(){while(cin>>n && n!=0){memset(send,false,sizeof(send));cin>>m;for(int i=0;i<n;i++){cin>>name[i];for(int j=0;j<name[i].length();j++){name[i][j] = tolower(name[i][j]);     //大写转换小写 }}int cnt = n;sort(name,name+n,cmp);for(int i=0;i<m;i++){cin>>tmp;for(int j=0;j<tmp.length();j++){tmp[j] = tolower(tmp[j]);}if(binary())cnt--;}cout<<cnt<<endl;}return 0;}