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

 

題解：字典樹，把每一個字符串的字串都用字典樹保存下來，並且記錄下每一個子串的個數，但是要注意每個字符串可能有相同的字串，此時只加一次。例如ababc,ab只加一次

 

#include 
#include 
#include 

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; //該字符串中該子字符已經出現了 
		    p->num++;  //該字串個數加一 
		}
	}
}

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] != '';j++)
			{
				insert(s + j,i);
			}
		}
		int q;
		scanf(%d,&q);
		for(int i = 0;i < q;i++)
		{
			scanf(%s,s);
			printf(%d
,find(s));
		}
		del(root);
	}
	
	
	return 0;
 }