HDU 2846 Repository (字典樹 後綴建樹)

 

Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2932 Accepted Submission(s): 1116

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

Source 2009 Multi-University Training Contest 4 - Host by HDU



題目大意：p個字符串，q個關鍵詞，問有多少個字符串包含關鍵詞

題目分析：對於每個字符串，我們按照其後綴建立字典樹，建樹時需要加1個id，表示這個分支來自第幾個字符串，不然會重復計數，比如樣例的第三組查詢

#include 
#include 
char s[25];
int id;

struct node
{
    node *next[26];
    int cnt;
    int id;
    node()
    {
        memset(next, NULL, sizeof(next));
        cnt = 0;
        id = -1;
    }
};

void Insert(node *p, char *s, int index, int id)
{
    for(int i = index; s[i] != ''; i++)
    {
        int idx = s[i] - 'a';
        if(p -> next[idx] == NULL)
            p -> next[idx] = new node();
        p = p -> next[idx];
        if(p -> id != id)
        {
            p -> id = id;
            p -> cnt ++;
        }
    }
}

int Search(node *p, char *s)
{
    for(int i = 0; s[i] != ''; i++)
    {
        int idx = s[i] - 'a';
        if(p -> next[idx] == NULL)
            return 0;
        p = p -> next[idx];
    }
    return p -> cnt;
}

int main()
{
    int p, q;
    id = 0;
    node *root = new node();
    scanf(%d, &p);
    for(int i = 0; i < p; i++)
    {
        scanf(%s, s);
        int len = strlen(s);
        for(int j = 0; j < len; j++)
            Insert(root, s, j, i);
    }
    scanf(%d, &q);
    for(int i = 0; i < q; i++)
    {
        scanf(%s, s);
        printf(%d
, Search(root, s));
    }
}