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

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
Your output should contain all the hat’s words, one per line, in alphabetical order.

[align=left]Sample Input[/align]

a
ahat
hat
hatword
hziee
word

[align=left]Sample Output[/align]

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.