NOJ 1121 Message Flood (Trie树 或者 map)

 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.

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

5 3
Inkfish
Henry
Carp
Max
Jericho
Carp
Max
Carp
0

3

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;}
}