noj 1121 Message Flood

Message Flood

时间限制(普通/Java) : 2000 MS/ 6000 MS          运行内存限制 : 65536 KByte
总提交 : 404            测试通过 : 108 

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

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

题目链接：http://acm.njupt.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1121

题目大意：有n个名字不重复的朋友发来祝福消息，有m条消息已回复给对应名字的朋友，求还需要回复多少个朋友。

解题思路：字典树记录n个名字，一个完整名字插入后在最后打上标记，对应名字查找到时记录查到个数，并把标记取消，避免重复查找。最后n减去已回复的朋友个数，即为还要回复的朋友个数。

代码如下：

#include <cstdio>#include <cstring>const int maxn=20005;char s[maxn];int ans;struct node{    bool end; node* next[26];node()   {        memset(next,NULL,sizeof(next));end=false;   }};void Insert(node *p,char *s){    for(int i=0;s[i]!='\0';i++)    {        int a;     //把名字的每个字母转换成数字被当做下标记录if(s[i] <= 'Z')     //不用区分大小写时同一个字母的大小写换乘同一个数字            a = s[i] - 'A';        else            a = s[i] - 'a';        if(p->next[a]==NULL)    //p的下一个下标为a的节点为空时，插入该节点            p->next[a]=new node();        p=p->next[a];     //p指针指向下一个节点    }    p->end=true;     //名字插入完成时名字末尾字母节点的end标记为true}void Search(node *p, char *s){    for(int i=0;s[i]!='\0';i++)    {int a;  //把名字的每个字母转换成数字被当做下标记录if(s[i]<='Z')    //不用区分大小写时同一个字母的大小写换乘同一个数字a=s[i]-'A';elsea=s[i]-'a';        if(p->next[a]==NULL)  //查找失败，返回            return ;        p=p->next[a];   //查找下一个节点    }if(p->end)  //找到，个数增加ans++;    p->end=false;  //为避免重复查找，标记为false，表示该串字符已不能作为名字被查找}int main(){    int n,m;    while(true)    {        ans=0;        node *root=new node();  //获得根节点scanf("%d",&n);if(n==0)break;        scanf("%d",&m);int nn=n;        while(nn--)        {            scanf("%s",s);            Insert(root ,s);   //从跟节点开始插入名字        }        while(m--)        {            scanf("%s",s);            Search(root,s);    //从根节点开始查找名字        }        printf("%d\n",n-ans);    }    return 0;}