HDU 1247 Hat’s Words （字典树 && map）

Posted 2020-09-10 mfmdaoyou

分析：一開始是用递归做的，没做出来。于是就换了如今的数组。即，把每个输入的字符串都存入二维数组中，然后创建字典树。输入和创建完成后，開始查找。

事实上一開始就读错题目了，题目要求字符串是由其它两个输入的字符串组成的前后缀，自己根本没有推断前缀是否满足。就直接推断后缀，一直想不通自己哪里错了，非常羞愧，水平还是不行。

查找分为前缀查找和后缀查找。事实上两个函数基本差点儿相同的。以下放代码。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

struct trie
{
     trie *next[26];
	 int v;    //字符同样的个数
	 trie()
	 {
		 memset(next,NULL,sizeof(next));
		 v=1;
	 }
};

trie *root=new trie();
char s[50001][101];

void creat_trie(char *str)
{
    int i,id;
	trie *p;
    for(p = root,i=0;str[i]; ++i)
    {
        id = str[i]-'a';
        if(p->next[id] == NULL)
        {
            p->next[id] = new trie();        
		}
		p = p->next[id];
	}
	p->v = -1;
}

int find_trie(char *str)
{
	int i=0,j,id;
	trie *p = root;
    for(;*str != '\0';)
    {
        id= *str - 'a' ;
		if (p->next[id] != NULL)
		{
			p = p->next[id];
			if(p->v == -1 && *(str+1) == '\0')
				return 1;
			str++;
		}
		else
			return 0;
    }
	return 0;

}

int find(char *str)
{
	trie *p = root;
	int m;
	for (;*str != '\0';)
	{
		m = *str - 'a';
		p = p->next[m];
		if(p != NULL)
		{
			if (p->v == -1 && find_trie(str+1))
			{
				return 1;
			}
			str++;
		}
		else 
			return 0;
		
	} 
	return 0;
}

int main()
{
	int N,n,i=0,j,t,m,flag=0;
	int a,b,c,d,k;
	int sum;
	trie *p;
	while (gets(s[i]),s[i][0] != '\0')//,s[i][0] != '\0'
	{		
		creat_trie(s[i++]);
	}
	for (j=0;j<i;j++)
	{
		   if (find(s[j]))
		   {
			   puts(s[j]);
		   }
	}
	return 0;
}

代码2：map容器

#include <iostream>
#include <map>
#include <cstring>
using namespace std;

map <string,int> m;
string s[50005];

int main() 
{
	int k=-1;
	while(cin>>s[++k])
	{
		m[s[k]] = 1;
	}
	for(int i=0;i<=k;i++)
	{
		int len = s[i].length();
		for(int j=1;j<len;j++)
		{
			string s1(s[i],0,j);  //从0開始的j个数
			string s2(s[i],j);    //从j開始(不包含)一直到结尾
			if(m[s1] == 1 && m[s2] == 1)
			{
				cout<<s[i]<<endl;
				break;
			}   
			
		}
	}
	return 0;
}

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9436    Accepted Submission(s): 3369

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