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

这题是有没有字符串是由其他两个字符串组成的
WA了很久的一个点 就是在枚举分割点，找到一个符合条件的，应该break 不然可能会重复输出一个字符串多次

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
using namespace std;
string s[50005];
vector<string> res;
bool cmp(string a,string b){
    return a<b;
}
struct Node{
    int cnt;
    Node* Next[26];
    Node(){
        cnt=0;
        for(int i=0;i<26;i++){
            Next[i]=NULL;
        }
    }
};
Node* root;
void insert(string s){
    Node* p=root;
    int val;
    int l=s.size();
    for(int i=0;i<l;i++){
        val=s[i]-'a';
        if(p->Next[val]==NULL){
            p->Next[val]=new Node();
        }
        p=p->Next[val];
    }
    p->cnt++;
}
bool find(string s){
    int l=s.size();
    Node* p=root;
    int val;
    for(int i=0;i<l;i++){
        val=s[i]-'a';
        if(p->Next[val]==NULL){
            return false;
        }
        p=p->Next[val];
    }
    if(p->cnt>0){
        return true;
    }
    else{
        return false;
    }
}
int main(void){
    //freopen("data.txt","r",stdin);
    int c=0;
    root=new Node();
    while(cin >> s[c]){
        insert(s[c]);
        c++;
    }
    for(int i=0;i<c;i++){
        //枚举分解点
        int len=s[i].size();
        for(int k=1;k<len;k++){
            string s1=s[i].substr(0,k);
            string s2=s[i].substr(k,len-k+1);
            //cout << s1 <<" "<<s2 << endl;
            if(find(s1) && find(s2)){
                res.push_back(s[i]);
                break;
            }
        }
    }
    sort(res.begin(),res.end(),cmp);
    int l=res.size();
    for(int i=0;i<l;i++){
        cout << res[i] << endl;
    }
    return 0;
}