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

題意：給定P個字符串，然後有Q個詢問，每次詢問給出一個字符串，問給定的P個串中有多少個串包含了它。
解析：可以利用字典樹處理。把每個給出的串的所有子串都建立到字典樹中去 ，注意的是，同一個單詞中相同的子串不重復計算。
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2846
代碼清單：

#include
#include
#include
#include
using namespace std;
const int MAX = 26;

struct trie{
    int point;   //計數
    int first;   //標記判斷是否是同一個單詞
    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;inext[pos]==NULL){
                q=new trie;
                q->point=1;
                q->first=pose;
                for(int k=0;knext[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;inext[pos]==NULL)
            return 0;
        p=p->next[pos];
    }
    return p->point;
}

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

int main(){
    for(int i=0;inext[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;
}