Hdu1247hat's words (combining words, Dictionary tree + DFS)

Problem descriptiona hat's word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary. You are trying to find all the hat's words in a dictionary.
Inputstandard 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.
Outputyour output shoshould contain all the hat's words, one per line, in alphabetical order.
Sample Input

aahathathatwordhzieeword

Sample output

ahathatword
#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;typedef struct nn{    int flag;    struct nn *next[26];}node;typedef struct strin{    int len,indx;    char str[105];}String;int cmp(String s1,String s2){    return s1.len<s2.len;}int cmp1(String s1,String s2){    return s1.indx<s2.indx;}node *builde(){    node *p=new node;    p->flag=0;    for(int i=0;i<26;i++)    p->next[i]=NULL;    return p;}node *root;void insert(char s[]){    node *p=root;    for(int i=0;s[i]!='\0';i++)    {        if(p->next[s[i]-'a']==NULL)        p->next[s[i]-'a']=builde();        p=p->next[s[i]-'a'];    }    p->flag=1;}String s[50005];int flog;void dfs(int k,int si,int num){    node *p=root;    if(num>2)         return ;    if(si==s[k].len)    {        if(num==2)flog=1; return ;    }    for(int j=si;j<s[k].len;j++)    {        if(p->next[s[k].str[j]-'a']==NULL)            return ;        p=p->next[s[k].str[j]-'a'];        if(p->flag)        {            dfs(k,j+1,num+1); if(flog!=0) return ;        }    }}int main(){    int n=0,m=0;    while(scanf("%s",s[n].str)>0&&s[n].str[0]!='#')    {        s[n].len=strlen(s[n].str); s[n].indx=n; n++;    }    sort(s,s+n,cmp);    root=builde();    for(int i=0;i<n;i++)    {       flog=0; dfs(i,0,0);       if(flog==1)s[m++]=s[i];       insert(s[i].str);    }    sort(s,s+m,cmp1);    for(int i=0;i<m;i++)    printf("%s\n",s[i].str);}