hdu 1247 Hat’s Words（字典树+分段判断）-优快云博客

本文介绍了一种使用字典树和指针链表的方法来找出所有由两个其他单词拼接而成的Hat's Words。通过枚举每个单词的所有可能分割，判断分割后的两部分是否分别存在于字典中。

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Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2391 Accepted Submission(s): 873

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
 

题目大意：给出你很多单词，求哪些单词是由这里其他单词（2个）连接组成的，注意！2个相同的单词连接也可以的！

又是字典树，不过要操作一下指针链表。方法是：将每一个单词分成两部分，分成两部分有len-1种方法，对每次情况进行枚举判断，判断是否组成已有的单词。这题也可以用stl的映射来做的，那个代码很短，也还好吧。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1247

代码：

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <string>
using namespace std;

char sh[50005][20];

struct node
{
    bool flag;
    node *next[26];
};

node *root, memory[5000005];
int cnt = 0;

node *create()
{
    node *p = &memory[cnt++];
    int i;
    for(i = 0; i < 26; i++)
    {
        p->next[i] = NULL;
    }
    p->flag = false;
    return p;
}

void insert(char *s)
{
    node *p = root;
    int i, k;
    for(i = 0; s[i]; i++)
    {
        k = s[i] - 'a';
        if(p->next[k] == NULL)
        {
            p->next[k] = create();
        }
        p = p->next[k];
    }
    p->flag = true;
}

bool search(char *s)
{
    int i, j, k;
    bool tar;
    for(i = 0; s[i]; i++)
    {
        node *p = root;
        for(j = 0; j < i; j++)
        {
            k = s[j] - 'a';
            p = p->next[k];
        }
        if(p->flag == 1)
        {
            node *q = root;
            tar = true;
            for(j = i; s[j]; j++)
            {
                k = s[j] - 'a';
                if(q->next[k] == NULL)
                {
                    tar = false;
                    break;
                }
                q = q->next[k];
            }
            if(q->flag == 1 && tar)
            {
                return true;
            }
        }
    }

    return false;
}

int main()
{
    int i, n;
    root = create();
    i = 0;
    while(scanf("%s", sh[i]) != EOF)
    {
        insert(sh[i]);
        i++;
    }
    n = i;
    for(i = 0; i < n; i++)
    {
        if(search(sh[i]))
        {
            printf("%s\n", sh[i]);
        }
    }

    return 0;
}