HDU【1247】Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11687 Accepted Submission(s): 4165



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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
const int maxnode = 5000000;
const int sigma_size = 26;
char str1[100],str2[100];
char s[50005][100];
struct Trie
{
int ch[maxnode][sigma_size];
int val[maxnode];
int sz;
Trie()
{
sz = 1;
memset(ch[0],0,sizeof(ch[0]));
}
int idx(char c)
{
return c - 'a';
}
void Insert(char ss[100])
{
int u = 0 ,n = strlen(ss);
for(int i = 0 ; i < n ; i++)
{
int c = idx(ss[i]);
if(!ch[u][c])
{
memset(ch[sz],0,sizeof(ch[sz]));
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = 1;
}
int Inquire(char sa[100])
{
int u = 0;
for(int i = 0;i < strlen(sa);i++)
{
int c = idx(sa[i]);
if(ch[u][c])
u = ch[u][c];
else return 0;
}
if(val[u]) return 1;
else return 0;
}
void tackle(char s[50005][100],int n)
{
int i,j;
for(i = 0 ;i < n ; i++)
{
int m = strlen(s[i]);
if(m != 1)
for(j = 1; j < strlen(s[i]); j++)
{
memset(str1,'\0',sizeof(str1));
memset(str2,'\0',sizeof(str2));
strncpy(str1,s[i],j);          //对每个字符串进行分解，然后判断这个字符串是否在字典树中
strcpy(str2,s[i]+j);
if(Inquire(str1) && Inquire(str2))                 {
printf("%s\n",s[i]);
break;
}
}
}
}
};
Trie tree;
int main()
{
int i = 0;
while(scanf("%s",s[i])!=EOF)
{
tree.Insert(s[i]);
i++;
}
tree.tackle(s,i);
return 0;
}