Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 5167    Accepted Submission(s): 1764

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

#include <bits/stdc++.h> using namespace std; #define mst(a,b) memset((a),(b),sizeof(a)) #define f(i,a,b) for(int i=(a);i<=(b);++i) const int maxn = 26; const int mod = 1e9+7; const int INF = 1e9; #define ll long long #define rush() int T;scanf("%d",&T);while(T--) struct Trie { int id; int cnt; int next[maxn]; }; int ptr=0; Trie word[1000001]; void Init(int z) { for(int i=0; i<maxn; i++) { word[z].next[i]=-1; word[z].cnt=0; } } void Insert(char*s,int len,int x) { int now=0; for(int i=0; i<len; i++) { int t=s[i]-'a'; if(word[now].next[t]==-1) { word[now].next[t]=++ptr; Init(ptr); now=ptr; } else now=word[now].next[t]; if(word[now].id!=x) //避免形如add造成的问题 { word[now].cnt++; word[now].id=x; } } } int find(char*s) { int now=0; int len=strlen(s); for(int i=0; i<len; i++) { int t=s[i]-'a'; if(word[now].next[t]==-1) return 0; now=word[now].next[t]; } return word[now].cnt; } int main() { int n,m,len; char s[21]; Init(0); scanf("%d",&n); while(n--) { scanf("%s",s); len=strlen(s); for(int i=0; i<len; i++) Insert(s+i,len-i,n+1); } scanf("%d",&m); while(m--) { scanf("%s",s); printf("%d\n",find(s)); } return 0; }