题意：给定单词集合S，包含若干单词，找出S中所有满足这样条件的元素p：p==str1+str2 && str1属于S &&　str2属于S
解法：暴力搜；或者用set的查找函数

You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 120,000 words.

Output

Your output should contain all the compound words, one per line, in alphabetical order.

Sample Input

a alien born less lien never nevertheless new newborn the zebra

 

Sample Output

alien newborn


set的搜索：
#include<iostream>
#include<set>
using namespace std;

int main() {
    set<string> s;
    string tmp;
    while (cin >> tmp)
        s.insert(tmp);
    set<string>::iterator it = s.begin();
    for (it; it != s.end(); it++) {
        tmp = *it;
        for (int i = 1; i < tmp.length(); i++) {
            if (s.find(tmp.substr(0, i)) != s.end() && s.find(tmp.substr(i, tmp.length() - i)) != s.end()) {
                cout << tmp << endl;
                break;
            }
        }
    }
    return 0;
}

暴力搜：
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int main(){
    //    freopen("in.txt","r",stdin);
    vector<string>vs[26];
    string s;
    while(getline(cin,s))
        if(s.length())vs[s[0]-'a'].push_back(s);
    for(int i=0; i<26; i++){
        if(vs[i].size()>=2){
            for(int j=1; j<vs[i].size(); j++){
                bool judge = true;
                for(int k=0; k<j; k++){
                    if(vs[i][j].find(vs[i][k])==0){
                        string t = vs[i][j].substr(vs[i][k].length());
                        if(!vs[t[0]-'a'].empty()){
                            for(int z=0; z<vs[t[0]-'a'].size(); z++){
                                if(vs[t[0]-'a'][z]==t){
                                    if(judge)cout << vs[i][j] << endl;
                                    judge = false;
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    return 0;
}


找到单词后注意标记，保证只输出1次。