?
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
题目大意：p个字符串，q个关键词，问有多少个字符串包含关键词
题目分析：对于每个字符串，我们按照其后缀建立字典树，建树时需要加1个id，表示这个分支来自第几个字符串，不然会重复计数，比如样例的第三组查询
#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)); } }
?