HDOJ Hat’s Words 1247【经典字典树】

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9643    Accepted Submission(s): 3446

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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建树。

拆分单词  然后在字典树里查找拆分后的两个单词

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1000000+1000
using namespace std;
int c[MAX][27];
bool v[MAX];
int tot;
char st[50100][100];
void init()
{
	tot=1;
	memset(c[0],0,sizeof(c[0]));
	memset(v,false,sizeof(v));
}
int insert(char *s)
{
	int l=strlen(s);
	int u=0;
	for(int i=0;i<l;i++)
	{
		int temp=s[i]-'a';
		if(!c[u][temp])
		{
			memset(c[tot],0,sizeof(c[tot]));
			v[tot]=0;
			c[u][temp]=tot++;
		}
		u=c[u][temp];
	}
	v[u]=1;  //记录每个单词最后一个字母。 
}
bool search(char *s)
{
	int l=strlen(s);
	int u=0;
	for(int i=0;i<l;i++)
	{
		int temp=s[i]-'a';
		if(!c[u][temp])
			return false;
		u=c[u][temp];
	}
	return v[u];
}
int main()
{
	int cnt=0;
	init();
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
	while(scanf("%s",st[cnt])!=EOF)
	{
		insert(st[cnt]);
		cnt++;
	}
	for(int i=0;i<cnt;i++)
	{
		int l=strlen(st[i]);
		char t1[100]={'\0'};
		char t2[100]={'\0'};
		for(int j=1;j<l-1;j++) //最后一个字母不用copy了 因为那就相当于找此整个单词 
		{
			strncpy(t1,st[i],j);
			strncpy(t2,st[i]+j,l-j+1); //记得把\0也copy过去。 
			if(search(t1)&&search(t2))
			{
			
				printf("%s\n",st[i]);
				break;
			}
		}
	}
	return 0;
}