Hat’s Words

Hat’s Words

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

Total Submission(s): 5620    Accepted Submission(s): 2091

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

 

字典树题目，WA了，10次，好多细节要考虑。哎，学习了。ACM要靠经验哦。

#include <iostream>
#include <string>
#include <vector>
using namespace std;
vector<string> hat;
struct dictree
{
	dictree *child[26];
	bool final;
	dictree()
	{
		memset(child,0,sizeof(child));
		final=false;
	}
};
dictree *root;
void insert(string source)
{
	dictree *current,*newnode;
	int len = source.length();
	if(!len)
		return ;
	current = root;
	for(int i=0;i<len;i++)
	{
		if(current->child[source[i]-'a']==0)
		{
			newnode = new dictree();
			current->child[source[i]-'a']=newnode;
		}
			current=current->child[source[i]-'a'];
	}
	current->final=true;
}
bool find(string source)
{
	dictree *current=root;
	for(int i=0;i<(int)source.size();i++)
	{
		if(current->child[source[i]-'a']!=0)
			current = current->child[source[i]-'a'];
		else
			return false;/*注意这个判断，WA了好几次的罪魁祸首啊。因为若少此一句，如果在某个终止结点处，source还没遍历完但是下一个不匹配。就会跳过这个字符去匹配下一个，好坑啊*/
	}
	return current->final;
}
int main()
{
	string temp;
	root=new dictree();
	while(cin>>temp)
	{
		insert(temp);
		hat.push_back(temp);
	}
	for(int j=0;j<(int)hat.size();j++)
	{
		string t=hat[j];
		unsigned len=t.size();
		for(int i=1;i!=len;i++)
		{
			string s1(t,0,i);
			string s2(t,i);
			if(find(s1)&&find(s2))
			{
					cout<<hat[j]<<endl;
					break;/*当满足条件是立刻跳出，否则可能会多次输出同一个
			}
		}
	}
	return 0;
}