HDU 1247 Hat’s Words（字典树活用）

Hat’s Words
Time Limit : 2000 / 1000 MS(Java / Others)    Memory Limit : 65536 / 32768 K(Java / Others)
Total Submission(s) : 18969    Accepted Submission(s) : 6689

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
a
ahat
hat
hatword
hziee
word

Sample Output
ahat
hatword

题意：按字典序输入几个单词，按字典序输出由另外两个单词组成的单词

分析：先建树，后查询每个单词，如果查询到第一个原有单词再接着查询剩余部分，剩余部分也是原有单词组成则输出该单词

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
char temp[50001][20];
bool vis[300001];
int t[300001][30],pos=1,num[300001];
void insert(char *s)//建树
{
	int rt = 0;
	int len = strlen(s);
	for (int i = 0; i < len; i++)
	{
		int x = s[i] - 'a';
		if (!t[rt][x])
			t[rt][x] = pos++;
		rt = t[rt][x];
	}
	vis[rt] = 1;//标记
}
bool search1(char *s)//查询后部分是否是已有单词构成
{
	int rt = 0;
	for (int i = 0; s[i]; i++)
	{
		int x = s[i] - 'a';
		if (!t[rt][x])
			return 0;
		rt = t[rt][x];
	}
	if (vis[rt])//验证尾结点是否为查到的单词的尾结点
		return 1;
	else
		return 0;
}
bool search(char *s)//前部分
{
	int rt = 0;
	for (int i = 0; s[i]; i++)
	{
		int x = s[i] - 'a';
		if (vis[rt] && search1(s + i))//rt为前部分单词的尾结点，验证后部分是否为已有单词
			return 1;
		rt = t[rt][x];
	}
	return 0;
}
int main()
{
	int i = 0;
	while (~scanf("%s", temp[i]))
		insert(temp[i++]);
	for (int j = 0; j <i; j++)
	{
		//cout << temp[j] << endl;
		if (search(temp[j]))
			printf("%s
", temp[j]);
	}
	return 0;
}