hdu 1247 Hat’s Words（字典树+分段判断）

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;
}