HDU 1247 Trie树

题意：按字典序输入一系列单词，输入结束后，判断哪些单词是由其余两个单词连接而成。若是，则输出（输出也按字典序）。

#include <iostream>
using namespace std;

char list[50001][26];
const int kind = 26;
int cnt = 0;

struct Treenode
{
	bool flag;
	Treenode *next[kind];
	Treenode()
	{
		flag = false;
		for ( int i = 0; i < kind; ++i )
			next[i] = NULL;
	}
} node[100000];

void Insert ( Treenode *&root, char *word )
{
	Treenode *location = root;
	if ( location == NULL )
	{
		location = &node[cnt++];
		root = location;
	}

	int id, i = 0;
	while ( word[i] )
	{
		id = word[i] - 'a';
		if ( location->next[id] == NULL )
			location->next[id] = &node[cnt++];
		location = location->next[id];
		++i;
	}
	location->flag = true; /*注意标记单词的最后一个字母，说明在这之前的字母组成一个单词，以便查询时划分*/
}

bool check ( Treenode *root, char* word )
{
	Treenode *location = root;
	if ( location == NULL ) return false;

	int id, i = 0, len = strlen(word);
	while ( word[i] )
	{
		id = word[i] - 'a';
		if ( location->next[id] == NULL )
			return false;
		if ( i == len-1 && location->next[id]->flag == true )
			return true;/*查找到最后一个字符时，节点的标记应该是true,这样才是完全匹配的*/
		location = location->next[id];
		++i;
	}
	return false;
}

bool Search ( Treenode *root, char * word )
{
	Treenode *location = root;
	if ( location == NULL ) return false;

	int id, i = 0, len = strlen(word);
	while ( word[i] )
	{
		id = word[i] - 'a';
		if ( location->next[id] == NULL )
			return false;
		if ( location->next[id]->flag == true )/*当检测到一个标记时说明单词的前半部分已经检测到，现在只需检测后面一部分是否能由另一个单词构成*/
		{
			if ( i + 1 < len && check(root,&word[i+1]) )
				return true;
		}
		location = location->next[id];
		++i;
	}
	return false;
}
		
int main()
{
	int i = 0, j;
	Treenode *root = NULL;
	while ( scanf("%s", list[i]) != EOF )
	{
		Insert(root,list[i]);
		++i;
	}
	
	for ( j = 0; j < i; ++j )
	{
		if ( Search(root,list[j]) )
			printf("%s\n",list[j]);
	}
	return 0;
}