NOJ 1121 Message Flood (Trie树 或者 map)

Message Flood

时间限制(普通/Java):2000MS/6000MS         运行内存限制:65536KByte
总提交:399          测试通过:105

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 andcase 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

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

题目大意：一个人要传消息给他的n个朋友，其中已经有m个（可能重复）收到了，问这个人还要发多少消息

题目分析：如果用trie树做注意红色标出的意思是不分大小写，建立26叉字典树，基本建树插入，ans初始化为n，查找的时候找到一个将其从字典中删除即可，用set + string做比较费时，但是代码很好写，很好理解

Trie树：

#include <cstdio>#include <cstring>char s[11];int cnt, ans;int change(char ch){    if(ch <= 'Z' && ch >= 'A')        return ch - 'A';    if(ch <= 'z' && ch >= 'a')        return ch - 'a'; }struct node{       node *next[26];    bool end;    node()    {        memset(next, 0, sizeof(next));        end = false;    }};void Insert(node *p, char *s){    for(int i = 0; s[i] != '\0'; i++)    {        int idx = change(s[i]);        if(p -> next[idx] == NULL)            p -> next[idx] = new node();        p = p -> next[idx];    }    p -> end = true;}void Search(node *p, char *s){       int i;    for(i = 0; s[i] != '\0'; i++)    {        int idx = change(s[i]);        if(p -> next[idx] == NULL)            return;        p = p -> next[idx];    }    if(p -> end)    {        ans --;        p -> end = false;    }}int main(){    int n, m;    while(scanf("%d", &n) && n)    {        ans = n;        node *root = new node();        scanf("%d", &m);        for(int i = 0; i < n; i++)        {            scanf("%s", s);            Insert(root, s);        }        for(int i = 0; i < m; i++)        {            scanf("%s", s);            Search(root, s);        }           printf("%d\n", ans);    }}

set + string：

#include <cstdio>#include <string>#include <iostream>#include <set>using namespace std;int main(){    int n, m;    string str;    while(scanf("%d %d", &n, &m) != EOF && n)    {        set <string> s;        for(int i = 0; i < n; i++)        {            cin >> str;            for(int j = 0; j < str.length(); j++)                str[j] = toupper(str[j]);            s.insert(str);        }        for(int i = 0; i < m; i++)        {            cin >> str;            for(int j = 0; j < str.length(); j++)                str[j] = toupper(str[j]);            set <string> :: iterator it = s.find(str);            if(it != s.end())                s.erase(str);        }        cout << s.size() << endl;    }}