HDU 2846 Repository

最新推荐文章于 2020-01-28 22:01:54 发布

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最新推荐文章于 2020-01-28 22:01:54 发布

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分类专栏：字典树

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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

/*
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;
}