字典树 模板
http://acm.hdu.edu.cn/showproblem.php?pid=1251
建立字典树，求以……为前缀的单词个数
统计难题
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 3656 Accepted Submission(s): 1295
注意:本题只有一组测试数据,处理到文件结束.
banana band bee absolute acm ba b band abc
2 3 1 0
#include<cstdio> #include<cstring> using namespace std; int n,m,ans; char str[10]; struct dic_tree//构造字典树，26叉树，根结点为空 { dic_tree *node[26];//26个结点 int num;//record the num of child bool terminal;//表示该结点位置是否是单词的结束位 dic_tree()//构造函数,初始化终点情况和结点 { for(int i = 0;i < 26;++i) { node[i] = NULL;//初始化空结点,node[0]表示'a'在该单词中 terminal = 0; num=0; } } }*root;//树是通过指针形式来表达的 void build(dic_tree *p,char str[],int end)//生成字典树的函数 { int ix = 0; p->num++; while(ix < end)//循环直到读完单词最后一个字母 { int c = str[ix] - 'a'; if(p->node[c])//如果到node[c]连通 { p = p->node[c]; if(ix == end - 1) p->terminal = 1; ++ix; } else { p->node[c] = new dic_tree; //如果不连通，则在该结点生成一个26叉树 p = p->node[c]; if(ix == end - 1) p->terminal = 1; ++ix; } p->num++; } } int search(dic_tree *p,char str[],int end) { int ix = 0; while(ix < end) { int c = str[ix] - 'a'; if(p->node[c]) { p = p->node[c]; if(ix==end-1) { p->terminal = 0; return p->num; } ++ix; } else return 0; } } int main() { //freopen("in.txt","r",stdin); root=new dic_tree; while(gets(str)&&str[0]!='/0') { build(root,str,strlen(str));//生成字典树 } while(scanf("%s",str)!=EOF) printf("%d/n",search(root,str,strlen(str)));//搜索字典树 return 0; }
当然，这道题还可以用map，只是时间较长：(*^__^*)
#include<iostream> #include<map> #include<string> using namespace std; int main() { int i,j,k,len; string str;char temp[15],temp1[15]; map <string,int> mymap; while(gets(temp)) { if(temp[0]=='/n') break; len=strlen(temp); if(len==0) break; for(i=0;i<len;i++)//求出某个字符串的所有前缀,并用MAP存起来 { for(j=0;j<=i;j++) temp1[j]=temp[j];temp1[j]='/0'; str.assign(temp1); mymap[str]++; } } while(scanf("%s",&temp)!=EOF) cout<<mymap[temp]<<endl;//此时直接输出结果即可 return 0; }
http://acm.njupt.edu.cn/problemdetail.do?method=showdetail&id=1121
Message Flood
总提交:139 测试通过:27
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
//搜索+字典树(Trie) //维基百科关于字典树的介绍 //Trie，又称单词查找树，是一种树形结构，是一种哈希树的变种。 //典型应用是用于统计和排序大量的字符串（但不仅限于字符串），所以经常被搜索引擎系统用于文本词频统计。 //它的优点是：最大限度地减少无谓的字符串比较，查询效率比哈希表高。 #include<cstdio> #include<cstring> using namespace std; int n,m,ans; char str[10]; struct dic_tree//构造字典树，26叉树，根结点为空 { dic_tree *node[26];//26个结点 bool terminal;//表示该结点位置是否是单词的结束位 dic_tree()//构造函数,初始化终点情况和结点 { for(int i = 0;i < 26;++i) { node[i] = NULL;//初始化空结点,node[0]表示'a'在该单词中 terminal = 0; } } }*root;//树是通过指针形式来表达的 void build(dic_tree *p,char str[],int end)//生成字典树的函数 { int ix = 0; while(ix < end)//循环直到读完单词最后一个字母 { int c = str[ix] - 'a'; if(p->node[c])//如果到node[c]连通 { p = p->node[c]; if(ix == end - 1) p->terminal = 1; ++ix; } else { p->node[c] = new dic_tree; //如果不连通，则在该结点生成一个26叉树 p = p->node[c]; if(ix == end - 1) p->terminal = 1; ++ix; } } } void search(dic_tree *p,char str[],int end) { int ix = 0; while(ix < end) { int c = str[ix] - 'a'; if(p->node[c]) { p = p->node[c]; if(ix == end - 1 && p->terminal == 1) { --ans; p->terminal = 0; return; } ++ix; } else return; } } void release(dic_tree *p)//必须递归释放结点空间 { for(int i = 0;i < 26; i++) if(p->node[i]) release(p->node[i]); delete p; } void transcase(char str[]) { for(int i = 0;i < strlen(str);++i) { if(str[i] >= 'A' && str[i] <= 'Z') str[i] = str[i] - 'A' + 'a'; } }//转化大小写 int main() { //freopen("in.txt","r",stdin); while(scanf("%d",&n)) { if(n == 0) break; scanf("%d",&m); root = new dic_tree; ans = n; for(int i = 0;i < n;++i) { scanf("%s",str); transcase(str);//转换大小写 build(root,str,strlen(str));//生成字典树 } for(int i = 0;i < m;++i) { scanf("%s",str); transcase(str);//转换大小写 search(root,str,strlen(str));//搜索字典树 } printf("%d/n",ans); release(root);//记住构造树后一定要释放结点空间，否则会内存溢出 } return 0; }