Repository
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
/*
字典树 - 字符出现次数
判断仓库中字符串能作为某一字符串母串的数目
*/
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int ch[500010][26];
int val[500010];
int num[500010];
int sz;
void init()
{
sz = 1;
memset(ch, 0, sizeof(ch));
memset(val, 0, sizeof(val));
memset(num, -1, sizeof(num));
}
//将一个单词分为多部分存储,用num记录这些部分同属于第x个单词
void insert_s(string s, int x)
{
//add中存储add,dd与d,则算含d单词数目时仅加一
int u = 0, i, c, len = s.size();
for (i = 0; i < len; i++)
{
c = s[i] - 'a';
if (!ch[u][c])
{
ch[u][c] = sz++;
}
u = ch[u][c];
if (num[u] != x)
{
val[u]++;
num[u] = x;
}
}
}
int query(string s)
{
int u = 0, i, c, len = s.size();
for (i = 0; i < len; i++)
{
c = s[i] - 'a';
if (!ch[u][c])
{
return 0;
}
u = ch[u][c];
}
return val[u];
}
int main()
{
int p;
cin >> p;
init();
int k = 0;
while (p--)
{
string a;
cin >> a;
for (int i = 0; i < a.size(); i++)
{
insert_s(a.substr(i, a.size() - i), k);
}
k++;
}
int q;
cin >> q;
while (q--)
{
string a;
cin >> a;
cout << query(a) << endl;
}
return 0;
}