寒假训练--字典树--B - Compound Words

Problem E: Compound Words

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

aalienbornlesslienneverneverthelessnewnewbornthezebra

Sample Output

aliennewborn

寻找是有两个字符串组合成的字符串，并输出，先建树，再查找时，如果遇到flag= 1 ，就跳到另一个函数里 继续查找，  如果最后仍是flag= 1  那这个字符串就是由这两个flag= 1 的字符串组成  应该输出，   如果没找到，应该退回一开始的函数里，寻找下一个flag= 1 的点，继续查找

#include <stdio.h>#include <string.h>char s[120010][50] ;struct node{    int flag ;    node *next[26] ;};struct node *newnode(){    node *p = new node ;    p->flag = 0;    for(int i = 0 ; i < 26 ; i++)        p->next[i] = NULL ;    return p;}void gettree(node *root,char *s){    int i , l = strlen(s) ;    node *p = root ;    for(i = 0 ; i < l ; i++)    {        int k = s[i] - 'a' ;        if(p->next[k]==NULL)            p->next[k] = newnode();        p = p->next[k] ;    }    p->flag = 1 ;}int ff(node *root,char *s,int i, int l){    node *p = root ;    for( ; i < l ; i++)    {        int k = s[i] - 'a' ;        if(p->next[k] == NULL) return 0;        p = p->next[k] ;    }    if(p->flag)    {        printf("%s\n", s);        return 1;    }    return 0;}void f(node *root,char *s){    int i , l = strlen(s) ;    node *p = root ;    for(i = 0 ; i < l ; i++)    {        int k = s[i] - 'a' ;        if(p->next[k] == NULL) return ;        p = p->next[k] ;        if(p->flag)            if(ff(root,s,i+1,l))return;    }}int main(){    int i , n = 0 ;    node *head ;    while(scanf("%s", s[n++])!=EOF)    {    }    n--;    head = newnode();    for(i = 0 ; i < n ; i++)        gettree(head,s[i]);    for(i = 0 ; i < n ; i++)        f(head,s[i]);}