# Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1129    Accepted Submission(s): 382

Problem Description
When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.

Input
There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.

Output
For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

Sample Input
20
ae
af
ag
ah
ai
aj
ak
al
aes
5
b
a
d
s

Sample Output
0
20
11
11
2

/*hdu 2846 字典树的变形 2011.10.18*/ #include <iostream>#include<cstring> #define MAX 26using namespace std;typedef struct Trie_Node{    int count;                  //记录包含该结点的单词个数      int id;                     //最后一次经过此结点的商品ID     Trie_Node *next[MAX];}Trie;void insert(Trie *root,char *s,int id){    Trie *p=root;    while(*s!='\0')    {        if(p->next[*s-'a']==NULL)        {            Trie *temp=(Trie *)malloc(sizeof(Trie));            for(int i=0;i<MAX;i++)            {                temp->next[i]=NULL;            }            temp->count=0;            temp->id=-1;            //-1表示没有商品             p->next[*s-'a']=temp;        }        p=p->next[*s-'a'];        if(p->id!=id)             //如果当前结点的商品ID不等于要插入商品的ID，则计数器count++，并且重新置ID的值          {            p->id=id;            p->count++;        }        s++;    }}int search(Trie *root,char *s){    Trie *p=root;    for(int i=0;s[i]!='\0';i++)    {        if(p->next[s[i]-'a']==NULL)            return 0;        p=p->next[s[i]-'a'];    }        return p->count;}int main(int argc, char *argv[]){    int i,j;    int n,m;    char s[21];    Trie *root=(Trie *)malloc(sizeof(Trie));    for(i=0;i<MAX;i++)    {        root->next[i]=NULL;    }    root->count=0;    root->id=-1;    scanf("%d",&n);    for(i=0;i<n;i++)    {        scanf("%s",s);        for(j=0;j<strlen(s);j++)       //将字符串X=X1X2...Xn的分别以X1,X2...Xn开头的后缀字符串插入到Trie树中          {            insert(root,s+j,i);        }        }    scanf("%d",&m);    for(i=0;i<m;i++)    {        scanf("%s",s);        printf("%d\n",search(root,s));    }    return 0;}
尝试过用KMP去解决，但是查找复杂度至少为0(p*(len1+len2))，试着提交了一下，严重超时。
posted @ 2011-10-18 10:57  Matrix海子  阅读(1762)  评论(0编辑  收藏  举报