Message Flood

Message Flood

Time Limit: 1500MS Memory limit: 65536K

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

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 void strUpr( char *str )//转换小写为大写的函数，本题目这是关键点
 5 {
 6        int i, len = strlen(str);
 7
 8       for ( i = 0; i < len; i++ )
 9       {
10           if( str[i] <= 'z' && str[i] >= 'a' )
11               str[i] = str[i] - 'a' + 'A';
12       }
13 }
14 void sort(char a[][20],int l,int r)
15 {
16     char  x[20];
17     strcpy(x,a[l]);
18     int i=l,j=r;
19     if(l>=r)return ;
20     while(i<j)
21     {
22         while(i<j&&strcmp(a[j],x)>=0)j--;
23         strcpy(a[i],a[j]);
24         while(i<j&&strcmp(a[i],x)<=0)i++;
25         strcpy(a[j],a[i]);
26     }
27     strcpy(a[i],x);
28     sort(a,l,i-1);
29     sort(a,i+1,r);
30 }
31 int binsearch(char a[][20],int s,int t,char key[])
32 {
33     int low=s,high=t,mid;
34     if(s<=t)
35     {
36         mid=low+(high-low)/2;
37         if(strcmp(a[mid],key)==0)
38         return mid;
39         if(strcmp(a[mid],key)>0)
40         return binsearch(a,low,mid-1,key);
41         else
42         return binsearch(a,mid+1,high,key);
43     }
44     return -1;
45 }
46 int main()
47 {
48     int m;
49     while(scanf("%d",&m)&&m!=0)
50     {
51         char f[20001][20],g[20001][20];
52         if(m==0)break;
53         else
54         {
55             int sum=0;//sum1为了标记重复元素的个数
56             int n;
57             scanf("%d",&n);
58             int i;
59             for(i=0;i<=m-1;i++)
60             {
61                 scanf("%s",f[i]);
62                 strUpr(f[i]);
63             }
64             for(i=0;i<=n-1;i++)
65             {
66                 scanf("%s",g[i]);
67                 strUpr(g[i]);
68             }
69             sort(g,0,n-1);//快速排序
70             for(i=0;i<=m-2;i++)
71             {
72                     if(strcmp(f[i],f[i+1])!=0)//相邻元素不相同，去重操作
73                     {
74                         if(binsearch(g,0,n-1,f[i])!=-1)//找到
75                         sum=sum+1;
76                     }
77                      //else continue;//相邻元素相同的话，不操作，进入下一次循环
78             }
79 //补上最后一个元素的特判
80             if(binsearch(g,0,n-1,f[i])!=-1)//找到了的话
81                         sum=sum+1;
82             printf("%d\n",m-sum);
83         }
84     }
85     return 0;
86 }