hdu（2846）Repository

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

题解：字典树，把每个字符串的字串都用字典树保存下来，而且记录下每个子串的个数，可是要注意每个字符串可能有同样的字串。此时仅仅加一次。比如ababc,ab仅仅加一次

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

struct Node
{
	int id;
	int num;
	Node* next[26];
	Node()
	{
		id = -1;
		num = 0;
		for(int i = 0;i < 26;i++)
		{
			next[i] = NULL;
		}
	}
};

Node* root;

void insert(char* s,int k)
{
	int len = strlen(s);
	Node* p = root;
	for(int i = 0;i < len;i++)
	{
		if(p->next[s[i] - 'a'] == NULL)
		{
			Node* q = new Node();
			q->id = k;
			q->num = 1;
			p->next[s[i] - 'a'] = q;
		}
		p = p->next[s[i] - 'a'];
		if(p->id != k)
		{
			p->id = k; //该字符串中该子字符已经出现了
		    p->num++;  //该字串个数加一
		}
	}
}

int find(char* s)
{
	int len = strlen(s);
	Node* p = root;
	for(int i = 0;i < len;i++)
	{
		if(p->next[s[i] - 'a'] == NULL)
		{
			return 0;
		}
		p = p->next[s[i] - 'a'];
	}
	return p->num;
}

void del(Node*& p)
{
	for(int i = 0;i < 26;i++)
	{
		if(p->next[i] != NULL)
		{
			del(p->next[i]);
		}
	}
	delete p;
}

int main()
{
	int n;
	while(scanf("%d",&n) != EOF)
	{
		root = new Node();
		char s[100];
		for(int i = 0;i < n;i++)
		{
			scanf("%s",s);
			for(int j = 0;s[j] != '\0';j++)
			{
				insert(s + j,i);
			}
		}
		int q;
		scanf("%d",&q);
		for(int i = 0;i < q;i++)
		{
			scanf("%s",s);
			printf("%d\n",find(s));
		}
		del(root);
	}

	return 0;
 }