【hdu1247】Hat’s Words

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

Author

戴帽子的

题解

题目意思:给一堆串，判断哪个串可以又这堆串里的两个串相连得到

trie树。把串先读进去，然后枚举每种可能判断一下。

s.substr (i,j) 串s 从i开始截取j位

next要开27个。（0里边是空的）

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    int next[27];
    int w;
}t[50010];
string ss[50010];
string s;
int topt,cnt;
void add_trie()
{
    int l=s.length(),now = 0;
    for (int i=0;i<l;i++)
    {
        int x = s[i]-'a'+1;
        if (t[now].next[x])
            now = t[now].next[x];
        else 
        {
            t[now].next[x] = ++topt;
            now = topt;
        }
    }
    t[now].w=1;
}
int find()
{
    int l=s.length(),now = 0,p = 0;
    while (p<l)
    {
        if (!t[now].next[s[p]-'a'+1]) return 0;
        else
        {
            now = t[now].next[s[p]-'a'+1];
            p++;
        }
    }
    if (t[now].w) return 1;
        else return 0;
}
int main()
{
    while (cin>>ss[++cnt])
    {
        s = ss[cnt];
        add_trie();
    }
    for (int i=1;i<cnt;i++)
    {
        string s3=ss[i];
        int l=s3.length();
        for (int j=1;j<l;j++)
        {
            s=s3.substr(0,j);
            if (find())
            {
                s = s3.substr(j,l);
                if (find())
                {
                    cout<<s3<<endl;
                    break;
                }
            }
        }
    }    
}