Hat’s Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2414 Accepted Submission(s): 880
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
题意很清楚:就是给出若干个单词，找出其中能够由其它两个单词连接组合而成的单词。比如上面sample中的"ahat"能够由"a"和"hat"拼凑而成，"hatword"="hat"+"word".因此只需对每个单词进行切分，把切分得到的两个单词在原单词序列中查找，若能查到则符合条件。在这里为了降低查找的复杂度，采用Trie树存储原始单词序列。
实现代码:
/*hdu 1247 Hat's word 字典树 2011.10.16*/
#include <iostream>
#include<cstring>
#define MAX 26
using
namespace
std;
typedef
struct
Trie_Node
{
bool
isWord;
struct
Trie_Node *next[MAX];
}Trie;
char
s[50000][50];
void
insert(Trie *root,
char
*word)
//插入单词
{
Trie *p=root;
while
(*word!=
'\0'
)
{
if
(p->next[*word-
'a'
]==NULL)
{
Trie *temp=(Trie *)
malloc
(
sizeof
(Trie));
for
(
int
i=0;i<MAX;i++)
{
temp->next[i]=NULL;
}
temp->isWord=
false
;
p->next[*word-
'a'
]=temp;
}
p=p->next[*word-
'a'
];
word++;
}
p->isWord=
true
;
}
bool
search(Trie *root,
char
*word)
//查找单词是否存在
{
Trie *p=root;
for
(
int
i=0;word[i]!=
'\0'
;i++)
{
if
(p==NULL||p->next[word[i]-
'a'
]==NULL)
return
false
;
p=p->next[word[i]-
'a'
];
}
return
p->isWord;
}
void
del(Trie *root)
//释放空间
{
for
(
int
i=0;i<MAX;i++)
{
if
(root->next[i]!=NULL)
{
del(root->next[i]);
}
}
free
(root);
}
int
main(
int
argc,
char
*argv[])
{
int
i,j;
int
count=0;
char
str[50];
Trie *root=(Trie *)
malloc
(
sizeof
(Trie));
for
(i=0;i<MAX;i++)
{
root->next[i]=NULL;
}
root->isWord=
false
;
while
(
scanf
(
"%s"
,str)!=EOF)
{
strcpy
(s[count++],str);
insert(root,str);
}
for
(i=0;i<count;i++)
{
for
(j=1;j<=
strlen
(s[i])-1;j++)
{
char
temp1[50]={
'\0'
};
char
temp2[50]={
'\0'
};
strncpy
(temp1,s[i],j);
strncpy
(temp2,s[i]+j,
strlen
(s[i])-j);
if
(search(root,temp1)&&search(root,temp2))
{
printf
(
"%s\n"
,s[i]);
break
;
//注意查找成功之后，需跳出循环，否则可能会重复打印
}
}
}
del(root);
return
0;
}
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本文转载自海 子博客园博客，原文链接：http://www.cnblogs.com/dolphin0520/archive/2011/10/16/2214231.html
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