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:
给你一堆单词，让你输出其中的hat's单词(由两个单词拼接成)
简单字典树
将所有单词插入字典树后，对于每个单词枚举其字串，若左右两字串都出现的话是一个hat's 输出即可
注意
1.不要用构造函数清空节点，不要用构造函数清空节点，不要用构造函数清空节点！会内存超限（手动清空）
2.查找单词时注意判断单词该节点是否为一个单词
#include<iostream>
#include<string.h>
#include<string>
#include<stdio.h>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=500000+20;
struct Node
{
int cnt;
int net[26];
void clear()
{
cnt=0;
memset(net,-1,sizeof(net));
}
} node[maxn];//每个节点都代表一个路径
int top;
void clear_tree()
{
node[0].clear();
top=0;
}
int hash_letter(char c)
{
return c-'a';
}
void insert_node(char *str)
{
int now=0;
while(*str)
{
int i=hash_letter(*str);
if(node[now].net[i]==-1)
{
node[top+1].clear();
node[now].net[i]=++top;
}
now=node[now].net[i];
++str;
}
node[now].cnt=1;
}
int search_node(char *str)//查看是否有这个单词
{
//根据str返回前缀结束指针
int now=0;
while(*str)//根据当前节点和字符选择该字符对应的节点
{
now=node[now].net[hash_letter(*str)];
if(now==-1)
return 0;
++str;
}
if(node[now].cnt)
return 1;
return 0;
}
char words[50100][100];
int main()
{
int n=0;
clear_tree();
top=0;
while(~scanf("%s",words[n]))
insert_node(words[n++]);
for(int i=0;i<n;++i)
{
char w1[100];
for(char *p=words[i]+1;*p;++p)//枚举字串
{
strcpy(w1,words[i]);
w1[p-words[i]]=' ';
if(search_node(w1)&&search_node(p))
{
printf("%s
",words[i]);
break;
}
}
}
}