Hat’s Words
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19500    Accepted Submission(s): 6867


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
 

Author
戴帽子的
 

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這道題也是比較簡單的trie樹的題，開始覺得很懵逼，最後才發現其實暴力就可以

對於每個單詞列舉斷點，然後查前後是否都存在

我們來計算下時間複雜度，假設n個單詞，每個單詞長度最大為m

那麼插入O（nm）；

查詢時候O（nmm）；

總複雜度O（nm+m2n）；

這道題的n是50000，m假設是30，那麼很明顯可以過

上程式碼

 1 #include<cstdio>
 2 #include<algorithm>
 3
 
 #include<cstring>
 4 #define idx(i) (i-'a')
 5 #define N 50011
 6 using namespace std;
 7 char in[N][300];
 8 int cnt=1,pp;
 9 struct TRIE{int nxt[30],cnt;}tree[N*300];
10 inline int regist(){return cnt++;}
11 void insert(char *now)
12 {
13     int c,rt=0,len=strlen(now);
14     for(int i=0;i<len;i++)
15     {
16         c=idx(now[i]);
17         if(!tree[rt].nxt[c])
18             tree[rt].nxt[c]=regist();
19         rt=tree[rt].nxt[c];
20     }
21     tree[rt].cnt=1;
22 }
23 bool find(char *now,int st,int ed)
24 {
25     int rt=0;
26     for(int i=st-1;i<ed;i++)
27     {
28         if(!tree[rt].nxt[idx(now[i])])return 0;
29         rt=tree[rt].nxt[idx(now[i])];
30     }
31     return tree[rt].cnt;
32 }
33 int main()
34 {
35     while(scanf("%s",in[++pp]+1)!=EOF)insert(in[pp]+1);
36     for(int i=1;i<pp;i++)
37     {
38         int len=strlen(in[i]+1);
39         for(int j=1;j<=len;j++)
40             if(find(in[i]+1,1,j)&&find(in[i]+1,j+1,len)){printf("%s\n",in[i]+1);break;}
41     }
42     return 0;
43 }