5.2.1 Hat’s Words

Hat’s Words

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

Total Submission(s): 135 Accepted Submission(s): 61

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

思路：正解该用字典树啊！我怎么这么偷懒，不学新东西，直接用map就过了 TAT

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
using namespace std;

map<string,int> mp;
const int maxn=50100,maxl=20;
int n,l;
char s[maxn][maxl],s1[maxl],s2[maxl];

void close()
{
    exit(0);
}

void work()
{
}

void init ()
{
n=0;
   while (scanf("%s",s[n])!=EOF)
    {
        mp[s[n]]=n;
        n++;
    }
    for (int i=0;i<n;i++)
    {
        l=strlen(s[i]);
        for (int j=0;j<l-1;j++)
        {
            memset(s1,'\0',sizeof(s1));
            memset(s2,'\0',sizeof(s2));
            for (int k=0;k<=j;k++)
                s1[k]=s[i][k];
            s1[j+1]='\0';
            for (int k=j+1;k<l;k++)
                s2[k-j-1]=s[i][k];
            s2[l]='\0';
        //    cout<<s[i]<<": "<<s1<<" "<<s2<<'\n';
            if (mp.find(s1)!=mp.end() && mp.find(s2)!=mp.end())
            {
                puts(s[i]);
                break;
            }
        }
    }
}

int main ()
{
    init();
    work();
    close();
    return 0;
}