HDU1247--Hat’s Words【STL--map】

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8718    Accepted Submission(s): 3136

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

aahathathatwordhzieeword

 

Sample Output

ahathatword

 

题意：按字典序输入一系列单词，输入结束后，判断哪些单词是由其余两个单词连接而成。若是，则输出（输出也按字典序）。

map+string，第一次用string的构造函数，百度了许多，就了解点了皮毛，用string果然精简，好省事啊。

关于string的构造函数，不懂的朋友可以看这里：

http://www.aichengxu.com/view/14112

#include <cstdio>#include <iostream>#include <algorithm>#include <map>using namespace std;map <string,int> m;string str[50010];int main (){    int i=-1;    while(cin>>str[++i]){        m[str[i]]=1;    }    int n=i;    for(i=0;i<=n;++i){        int len=str[i].size()-1;        for(int j=1;j<len;++j){            string s1(str[i],0,j);//将字符串str中，起始位置为0，长度为j的部分赋值给s1            string s2(str[i],j);  //将字符串str中，位置j以后的部分（包括j）赋值给s2（这里的位置是字符串的的下标）            if(m[s1]==1 && m[s2]==1){                cout<<str[i]<<endl;                break;            }        }    }    return 0;}

今天好累，写完这道就睡。