Hdoj 2846

原题链接

描述

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.

输入

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.

输出

For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

样例输入

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

样例输出

0
20
11
11
2

思路

字典树。
刚开始我也不知道怎么做，后来听大佬的思路的。
将每个单词的后缀都存入字典树，标记好源于同一个单词的后缀，然后判断。虽然时间复杂度很悬的样子，但还是AC了。

代码

#include <bits/stdc++.h>
#define maxn 900000
using namespace std;

struct node
{
	int a[26];
	int b;
	int val;
};

vector<node> trie;
int tot, sum;

void create(char st[], int num)
{
	int len = strlen(st);
	int u = 0;
	for(int i = 0; i < len; i++)
	{
		int v = st[i] - 'a';
		if(trie[u].a[v] == 0)
		{
			trie[u].a[v] = ++tot;
			node pnew; 
			for(int j = 0; j < 27; j++) pnew.a[j] = 0; 
			pnew.b = -1; pnew.val = 0;
			trie.push_back(pnew);
		}
		u = trie[u].a[v];
		if(trie[u].b != num)
		{
			trie[u].b = num;
			trie[u].val++;
		}
	}
}

void find(char st[])
{
	int len = strlen(st);
	int u = 0, f = 1;
	for(int i = 0; i < len; i++)
		if(trie[u].a[st[i] - 'a'])
			u = trie[u].a[st[i] - 'a'];
		else {f = 0; break;}
	if(f) sum += trie[u].val;
	
}

int main()
{
	node pnew; 
	for(int j = 0; j < 27; j++) pnew.a[j] = 0; 
	pnew.b = -1; pnew.val = 0;
	trie.push_back(pnew);
	int n; scanf("%d", &n);
	while(n--)
	{
		char st[21];
		scanf("%s", st);
		int len = strlen(st);
		for(int i = 0; i < len; i++)
			create(st + i, n);
	}
	scanf("%d", &n);
	while(n--)
	{
		char st[21];
		scanf("%s", st);
		sum = 0;
		find(st);
		printf("%d\n", sum);
	}
	return 0;
}