HDU 1247 Hat’s Words（字典树变形）

题目链接：

pid=1247" target="_blank">http://acm.hdu.edu.cn/showproblem.php?

pid=1247

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

a
ahat
hat
hatword
hziee
word

 

Sample Output

ahat
hatword

题意：

寻找出单次表中一个单词是由另外两个单词组成的。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <malloc.h>
#include <iostream>
using namespace std;
#define MAXN 26
char str[50017][117];
typedef struct Trie
{
    int v;//依据须要变化
    Trie *next[MAXN];
    //next是表示每层有多少种类的数，假设仅仅是小写字母，则26就可以，
    //若改为大写和小写字母，则是52，若再加上数字，则是62了
}Trie;
Trie *root;

void createTrie(char *str)
{
    int len = strlen(str);
    Trie *p = root, *q;
    for(int i = 0; i < len; i++)
    {
        int id = str[i]-'a';
        if(p->next[id] == NULL)
        {
            q = (Trie *)malloc(sizeof(Trie));
            q->v = 1;//初始v==1
            for(int j = 0; j < MAXN; j++)
                q->next[j] = NULL;
            p->next[id] = q;
            p = p->next[id];
        }
        else
        {
        //    p->next[id]->v++;
            p = p->next[id];
        }
    }
     p->v = -1;//若为结尾，则将v改成-1表示
}

int findTrie(char *str)
{
    int len = strlen(str);
    Trie *p = root;
    for(int i = 0; i < len; i++)
    {
        int id = str[i]-'a';
        p = p->next[id];
        if(p == NULL) //若为空集，表示不存以此为前缀的串
            return 0;
     //   if(p->v == -1)   //字符集中已有串是此串的前缀
     //       return -1;
    }
    //return -1;   //此串是字符集中某串的前缀
    if(p->v == -1)//说明是全然匹配
        return -1;
    else
        return 0;
}
int dealTrie(Trie* T)
{
    //动态字典树，有时会超内存,这是就要记得释放空间了
    if(T==NULL)
        return 0;
    for(int i = 0; i < MAXN; i++)
    {
        if(T->next[i]!=NULL)
            dealTrie(T->next[i]);
    }
    free(T);
    return 0;
}
int main()
{
    root = (Trie *)malloc(sizeof(Trie));
    for(int i = 0; i < MAXN; i++)
        root->next[i] = NULL;
    int cont = 0;
    while(scanf("%s",str[cont])!=EOF)
    {
        createTrie(str[cont]);
        cont++;
    }
    char a[117], b[117];
    for(int i = 0; i < cont; i++)
    {
        int len = strlen(str[i]);
        for(int j = 1; j < len; j++)
        {
            memset(a,'\0',sizeof(a));
            memset(b,'\0',sizeof(b));
            strncpy(a,str[i],j);
            strncpy(b,str[i]+j,len-j);
            if(findTrie(a)==-1 && findTrie(b)==-1)
            {
                printf("%s\n",str[i]);
                break;
            }
        }
    }
    return 0;
}