HDU 1247 Hat's Words (map+string)

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10993    Accepted Submission(s): 3944

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

 

这道题小弱初看时， 好像没有什么思路。然而其实计算机的最大好处就在计算的快速。 所以直接暴力就行啦！ 现在一看到题老是想是否要用到什么高级的结构，什么技巧， 其实有些题， 废话少说，直接上暴力就行啦。

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

map<string,int> M;
string str[50005];

int main()
{
    //freopen("in.txt", "r", stdin);
    int k = -1;
    while(cin>>str[++k])
    M[str[k]] = 1;
    for(int i=0; i<=k; i++)
    {
        int len= str[i].size()-1;
        for(int j=1; j<len; j++)
        {
            string s1(str[i], 0, j);
            string s2(str[i], j);
            if(M[s1]==1&&M[s2]==1)
            {
                cout<<str[i]<<endl;
                break;
            }
        }
    }
    return 0;
}

本题技巧： string  s1(str[i], 0, j);把str[i]中的0~j的字符赋给s1. string s2(str[i], j)把str[i]中的j~末尾的字符赋给s2.