A - Hat’s Words
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
通过map存单词,在遍历每个单词,枚举单词的拆分,通过map的find函数来查找拆分出的两个单词,查找的时间复杂度为O(logn)
#include<iostream>
#include<stdlib.h>
#include<string>
#include<queue>
#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
map<string,int> mp;
map<string,int>::iterator it;
int main()
{
ios::sync_with_stdio(false);
int sum=0;
string a;
while(cin>>a)
{
mp[a]=1;
}
for(it=mp.begin();it!=mp.end();it++)
{
a=it->first;
for(int i=0;i<a.size()-1;i++)
{
string a1(a,0,i);
string a2(a,i);
if(mp.find(a1)!=mp.end()&&mp.find(a2)!=mp.end())
{
cout<<a<<endl;
break;
}
}
}
}
、