HDU-2846 Repository (Trie 字典树入门题)

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本文介绍了一种使用Trie树处理大量字符串子串查询的方法。针对1万个长度为20的单词，实现高效的子串查找算法，并能准确回答10万个查询请求。通过详细解释算法流程及代码实现，帮助读者理解如何构建Trie树并进行有效搜索。

Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5891 Accepted Submission(s): 1954

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

#include <bits/stdc++.h>
using namespace std;
#define maxn 500000
int ch[maxn][27], sz[maxn], tot;
map<int, int>mp;
void insert(char s[], int j){
	int n = strlen(s), k = 0, y;
	for(int i = j; i < n; ++i){
		y = s[i] - 'a';
		if(ch[k][y] == 0){
			ch[k][y] = ++tot;
		}
		k = ch[k][y];
		if(mp[k] == 0){     //没出现过才计算个数
			sz[k]++;
			mp[k] = 1;
		}
	}
}
int query(char s[]){
	int n = strlen(s), k = 0, y;
	for(int i = 0; i < n; ++i){
		y = s[i] - 'a';
		if(ch[k][y] == 0){
			return 0;
		}
		k = ch[k][y];
	}
	return sz[k];
}
int main(){
	int n, ans = 0;
	char s[22];
	memset(sz, 0, sizeof(sz));
	tot = 0;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%s", s);
		int x = strlen(s);
		mp.clear();     
		for(int j = 0; j < x; ++j){
			insert(s, j);
		}
	}
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i){
		scanf("%s", s);
		printf("%d\n", query(s));
	}
}

/*
题意：1e4个单词，每个单词长度为20。1e5次询问，回答给定单词是多少
个单词的子串。

思路：对所有原单词先建trie，每个单词的所有子串都要建进去，即
复杂度O(20*20*10000)。对于每次询问，注意一下坑点：像add这种单词，
d作为子串出现了两次，但统计答案时只能算一次，所以在插入时注意处理
这种情况，然后对于每个结点i记录一下sz[i],常规遍历询问就行了。
*/