HDU 1247 Hat's words（字典树Trie)

标签：acm   字典树   

解题思路：

判断给出的单词是否恰好由另外两个单词组成，用栈保存每个子字符串的节点，从这个节点出发判断剩下的字符串是否在字典树中即可。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <stack>
#include <set>
using namespace std;
const int MAXN = 50000 + 10;
typedef struct Trie_Node
{
	bool isword;
	struct Trie_Node *next[26];
}Trie;
void insert(Trie *root, char *word)
{
	Trie *p = root;
	int i = 0;
	while(word[i] != '\0')
	{
		if(p->next[word[i]-'a'] == NULL)
		{
			Trie *temp = new Trie;
			temp->isword = false;
			for(int j=0;j<26;j++)
				temp->next[j] = NULL;
			p->next[word[i]-'a'] = temp;
		}
		p = p->next[word[i]-'a'];
		i++;
	}
	p->isword = true;
}
bool Find(Trie *root, char *word)
{
	Trie *p = root;
	int i = 0;
	stack<int> s;while(!s.empty()) s.pop();
	while(word[i] != '\0')
	{
		if(p->next[word[i]-'a'] == NULL)
            return 0;
        p = p->next[word[i]-'a'];
        if(p->isword && word[i]) s.push(i+1);
        i++;
	}
	while(!s.empty())
    {
        bool ok = true;
        i = s.top(); s.pop();
        p = root;
        while(word[i])
        {
            if(p->next[word[i]-'a'] == NULL)
            {
                ok = false;
                break;
            }
            p = p->next[word[i]-'a'];
            i++;
        }
        if(ok && p->isword)
            return 1;
    }
	return 0;
}
char s[MAXN][27];
int main()
{
	int n = 0;
	Trie *root = new Trie;
	root->isword = false;
	for(int j=0;j<26;j++)
		root->next[j] = NULL;
	while(scanf("%s", s[n]) != EOF)
	{
		insert(root, s[n]);
		n++;
	}
	for(int i=0;i<n;i++)
	{
		if(Find(root, s[i]))
         printf("%s\n", s[i]);
	}
	return 0;
}

HDU 1247 Hat's words（字典树Trie)

标签：acm   字典树   

原文地址：http://blog.csdn.net/moguxiaozhe/article/details/43735881