HDU_2846_Repository（字典树）

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2664    Accepted Submission(s): 1045

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
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
20
11
11
2
   
   
   
   

 
题意：给定P个字符串，然后有Q个询问，每次询问给出一个字符串，问给定的P个串中有多少个串包含了它。
解析：可以利用字典树处理。把每个给出的串的所有子串都建立到字典树中去 ，注意的是，同一个单词中相同的子串不重复计算。
题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=2846
代码清单：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAX = 26;

struct trie{
    int point;   //计数
    int first;   //标记判断是否是同一个单词
    trie *next[MAX];

};

trie *root=new trie();

int p,n;
char s[MAX];

void createTrie(char *s,int pose){
    trie *p, *q;
    int len=strlen(s),pos;
    for(int i=0;i<len;i++){
        p=root;
        for(int j=i;j<len;j++){
            pos=s[j]-'a';
            if(p->next[pos]==NULL){
                q=new trie();
                q->point=1;
                q->first=pose;
                for(int k=0;k<MAX;k++)
                    q->next[k]=NULL;
                p->next[pos]=q;
                p=p->next[pos];
            }
            else if(p->next[pos]->first==pose){
                p=p->next[pos];
            }
            else{
                p->next[pos]->point++;
                p->next[pos]->first=pose;
                p=p->next[pos];
            }
        }
    }
}

int findTrie(char *s){
    trie *p=root;
    int len=strlen(s),pos;
    for(int i=0;i<len;i++){
        pos=s[i]-'a';
        if(p->next[pos]==NULL)
            return 0;
        p=p->next[pos];
    }
    return p->point;
}

void delTrie(trie *Root){
    for(int i;i<MAX;i++){
        if(Root->next[i]!=NULL)
            delTrie(Root->next[i]);
    }
    free(Root);
}

int main(){
    for(int i=0;i<MAX;i++)
        root->next[i]=NULL;
    scanf("%d",&p);
    for(int i=1;i<=p;i++){
        scanf("%s",s);
        createTrie(s,i);
    }
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%s",s);
        printf("%d\n",findTrie(s));
    }
    delTrie(root);
    return 0;
}