Message Flood
Time Limit: 1500ms Memory limit: 65536K 有疑問？點這裏^_^
題目描述
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.
輸入
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.
輸出
For each case, print one integer in one line which indicates the number of left friends he must send.
示例輸入
5 3 Inkfish Henry Carp Max Jericho Carp Max Carp 0
示例輸出
3
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct code\\定義結構體
{
int len;
struct code *next[26];
};
struct code * newnode()\\節點初始化
{
int i;
struct code *p=(struct code *)malloc(sizeof(struct code));
p->len=0;
for(i=0; i<26; i++)
p->next[i]=NULL;
return p;
}
void insert(struct code *root,char *s)\\插入並建樹
{
struct code *p=root;
int i,len=strlen(s),t;
for(i=0; i<len; i++)
{
if(s[i]>='a'&&s[i]<='z')
t=s[i]-'a';
else t=s[i]-'A';
if(p->next[t]==NULL)
p->next[t]=newnode();
p=p->next[t];
}
p->len=1;
}
int find(struct code *root,char*s)\\查找
{
struct code *p=root;
int i,t,len=strlen(s);
for(i=0; i<len; i++)
{
if(s[i]>='a'&&s[i]<='z')
t=s[i]-'a';
else t=s[i]-'A';
if(p->next[t]==NULL)
return 0;
p=p->next[t];
}
if(p->len)
{
len=0;
return 1;
}
return 0;
}
int dealtrie(struct code *root)\\刪除樹，防止內存佔用過多
{
int i;
if(root==NULL)
return 0;
for(i=0; i<26; i++)
{
if(root->next[i]!=NULL)
dealtrie((root->next[i]));
}
free(root);
return 0;
}
int main()
{
int i,m,n,sum;
char s1[20000][100],s[100];
while(~scanf("%d",&n),n)
{
scanf("%d",&m);
for(i=0; i<n; i++)
{
scanf("%s",s1[i]);
}
struct code *root=newnode();
for(i=0; i<m; i++)
{
scanf("%s",s);
insert(root,s);
}
sum=0;
for(i=0; i<n; i++)
{
if(find(root,s1[i]))
sum++;
}
printf("%d\n",n-sum);
dealtrie(root);
}
return 0;
}