HDU 1247 - Hat’s Words - [字典树水题]

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1247

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：

“帽子单词”是指，若字典中的某个词，它是由其他任意两个单词连接起来组成的，则称它为“帽子单词”。

现在以字典序给出字典中的所有单词（不超过 $5e4$ 个），让你求出全部“帽子单词”。

题解：

先把字典建成字典树，然后对于每个单词，暴力地分成两个子串查找即可。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=5e4+5;

namespace Trie
{
    const int SIZE=maxn*32;
    int sz;
    struct TrieNode{
        int ed;
        int nxt[26];
    }trie[SIZE];
    void init(){sz=1;}
    void insert(const string& s)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            int ch=s[i]-'a';
            if(!trie[p].nxt[ch]) trie[p].nxt[ch]=++sz;
            p=trie[p].nxt[ch];
        }
        trie[p].ed++;
    }
    int search(const string& s)
    {
        int p=1;
        for(int i=0;i<s.size();i++)
        {
            p=trie[p].nxt[s[i]-'a'];
            if(!p) return 0;
        }
        return trie[p].ed;
    }
};
int tot;
string s[maxn];
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    Trie::init();
    tot=0;
    while(cin>>s[++tot]) Trie::insert(s[tot]);
    for(int i=1;i<=tot;i++)
    {
        bool ok=0;
        for(int k=1;k<s[i].size();k++)
        {
            if(Trie::search(s[i].substr(0,k)) && Trie::search(s[i].substr(k,s[i].size()-k)))
            {
                ok=1;
                break;
            }
        }
        if(ok) cout<<s[i]<<'\n';
    }
}