字典树

初步学了下字典树，，，有好多地方都还是搞不清楚为什么这样写，，，不过可以基本的套用模板。

套用模板做了这个题：hdu1247http://acm.hdu.edu.cn/showproblem.php?pid=1247

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11265    Accepted Submission(s): 4020

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
  

题意就是：给出一些单词，找出其中能由其他两个单词连接而成的单词。“ahat” = “a” + “hat”。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>

using namespace std;

const int MAXN = 1000000;
char s[50005][101];

struct Node
{
    int flag;
    int cnt;
    struct Node * next[26];
} tree[MAXN];

int t = 0;
struct Node *creat()
{
    struct Node *p;
    p = &tree[t ++];
    p->cnt = 1;
    p->flag = 0;
    for(int i = 0; i < 26; i++)
        p->next[i] = NULL;
    return p;
}

void Insert(struct Node **root,char *ss)
{
    struct Node *p;
    if(!(p = *root))p = *root = creat();
    int i = 0,k;
    while(ss[i])
    {
        k = ss[i ++] - 'a';
        if(p->next[k])p->next[k]->cnt ++;
        else p->next[k] = creat();
        p = p->next[k];
    }
    p->flag = 1;
}

int Search(struct Node **root,char *ss)
{
    struct Node *p;
    if(!(p = *root))return 0;
    int i = 0,k;
    while(ss[i])
    {
        k = ss[i++] - 'a';
        if(!(p->next[k]))return 0;
        p = p->next[k];
    }
    return p->flag;
}

int main()
{
//    freopen("in.txt","r",stdin);
    char s1[101],s2[101];
    struct Node *root = NULL;
    int i = 0;
    while(~scanf("%s",s[i]))
    {
        Insert(&root,s[i]);
        i ++;
    }
    for(int j = 0;j < i;j++)
    {
        memset(s1,0,sizeof(s1));
        int l = strlen(s[j]);
        for(int k = 0;k < l;k++)
        {
            s1[k] = s[j][k];
            if(Search(&root,s1))
            {
                memset(s2,0,sizeof(s2));
                for(int q = k + 1;q < l;q ++)
                {
                    s2[q - k - 1] = s[j][q];
                }
                if(Search(&root,s2))
                {
                    puts(s[j]);
                    break;
                }
            }
        }
    }
    return 0;
}

字典树： http://www.cnblogs.com/pony1993/archive/2012/07/18/2596730.html