【trie树】查找由其他两个字符串组成的串 HDU1247

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本文介绍了一种特殊的字典词汇——Hat's Words，即由字典中两个不同单词精确拼接而成的新词。文章通过提供算法和代码示例，详细讲解了如何在一个给定的字典中识别所有此类特殊词汇，利用前缀树（Trie）进行高效查找。

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19117 Accepted Submission(s): 6750

Problem Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

Sample Input

ahat

hat

hatword

hziee

word

Sample Output

ahat

hatword

#include <bits/stdc++.h>
using namespace std;
const int mn = 50010, mm = 85;

char ch[mn][mm];
bool ed[mn * mm];
int trie[mn * mm][30];

int num;
void add(char ch[])
{
	int len = strlen(ch);
	int r = 0;
	for (int i = 0; i < len; i++)
	{
		int t = ch[i] - 'a';
		if (trie[r][t] == 0)
			trie[r][t] = ++num;
		r = trie[r][t];
	}
	ed[r] = 1;
}
bool ask(char ch[])
{
	int r = 0;
	int len = strlen(ch);
	for (int i = 0; i < len; i++)
	{
		int t = ch[i] - 'a';
		if (trie[r][t] == 0)
			return 0;
		r = trie[r][t];
	}
	if (ed[r] == 1)
		return 1;
	else
		return 0;
}

int main()
{
	int cas = 1;
	while (~scanf("%s", ch[cas]))
	{
		add(ch[cas]);
		cas++;
	}

	char s1[mm], s2[mm];
	for (int i = 1; i < cas; i++)
	{
		int len = strlen(ch[i]);
		for (int j = 1; j < len - 1; j++) // 枚举断点
		{
			strncpy(s1, ch[i], j);
			s1[j] = '\0';
			if (ask(s1) == 0)
				continue;

			strncpy(s2, ch[i] + j, len - j);
			s2[len - j] = '\0';
			if (ask(s2) == 1)
			{
				printf("%s\n", ch[i]);
				break;
			}
		}
	} // O(n * m * m)
	return 0;
}