Hat’s Words
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.
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.
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：给你一个字典，问里面那些单词可以由两个字典中单词拼接而成。
PS:我开始理解的题目是单词是要以两个出现过的单词拼接起来才算是hat's word，比如a,a,aa,aa=a+a，如果是a,aa就不行了。
但是最后发现只要是出现过的就行了，数目无所谓，而且单词相同也可以（那two other words是什么意思啊）
题解：字典树，发现有匹配的前缀再去判断剩下的一段是不是有匹配就行了。
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
node *next[26];
int id,num;
node()
{
memset(next,0,sizeof(next));
id=0;
num=0;
}
}*head;
char s[50005];
vector<string> mat,out;
void build(char *s,node *head,int id)
{
int len=strlen(s),k;
for(int i=0; i<len; ++i)
{
k=s[i]-'a';
if(head->next[k]==NULL)
head->next[k]=new node();
head=head->next[k];
}
head->id=id;
head->num++;
}
bool bfs(char *s,node *head)
{
int len=strlen(s),j,k,h;
node *p=head,*q;
for(int i=0; i<len; ++i,p=p->next[k])
{
k=s[i]-'a';
if(p->next[k]==NULL) return false;
if(p->next[k]->id)
{
q=head;
for(j=i+1; j<len; ++j,q=q->next[h])
{
h=s[j]-'a';
if(q->next[h]==NULL) break;
}
if(j==len&&q->id)
//if((q->id!=p->next[k]->id)||(q->num>1))
return true;
}
}
return false;
}
int main()
{
head=new node();
for(int i=1; ~scanf("%s",s); ++i)
{
build(s,head,i);
mat.push_back(string(s));
}
for(int i=0; i<mat.size(); ++i)
{
strcpy(s,mat[i].c_str());
if(bfs(s,head))
out.push_back(mat[i]);
}
sort(out.begin(),out.end());
for(int i=0; i<out.size(); ++i)
printf("%s\n",out[i].c_str());
return 0;
}