Noj 1121 Message Flood (trie tree or map)

Message Flood time limit (normal/java): 2000ms/6000ms running memory limit: 65536KByte
Total Submissions: 399 test pass:

Title Description

Well, how does the feel about mobile phone? Your answer would probably be something as that "It's so convenient and benefits people a lot". However, if you ask Merlin this question to the New year's Eve, he'll definitely answer "what a trouble! I had to keep my fingers moving on the phone the whole night, because I had 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 had sent message to him, and he doesn ' t want to send an Other message to bother them (Merlin are so polite, he always replies each message he receives immediately). So, the before he begins to the send messages, he needs to the figure of how many friends is left-to-be sent. Please write a him.
Here's something that should note. First, Merlin's friend list is not ordered, and each name is alphabetic strings andCase insensitive. These names is guaranteed to is not duplicated. Second, some senders may send more than one message to Merlin, therefore the sender list may be duplicated. Third, Merlin is known-many people, that's why some message senders was even not included on his friend list.

Input

There is multiple test cases. In each case, at the first line there was numbers n and M (1<=n, m<=20000), which is the number of friends a nd the number of messages he has received. And then there is n lines of alphabetic strings (the length of each would be is less than), indicating the names of Mer Lin ' s friends, one pre line. After that there is m lines of alphabetic string s, which is the names of message senders.
The input is terminated by n=0.

Output

For each case, print one integer in one line which indicates the number of the left friends he must send.

Sample input

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

Sample output

3

Source of the topic

The ninth session of Zhongshan University Program design contest pre-selection

Title Link: http://acm.njupt.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1121

The main idea: a person to send a message to his n friends, of which already have m (May repeat) received, ask this person how much information to send

Title Analysis: If you use trie tree to pay attention to the meaning of red mark is not divided into the case, the establishment of 26-Fork Dictionary tree, Basic achievements inserted, ans initialized to N, find a time to find a way to delete it from the dictionary, with set + string to do a comparison, but the code is very good to write, very good understanding

Trie Tree:

#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]! = ' + '; 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]! = ' + '; 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;    }}

Noj 1121 Message Flood (trie tree or map)