HDU - 1247 - Hat's Words（字典树）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15814    Accepted Submission(s): 5637

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：给一组单词，找出可以分为两部分且两部分单词均在该组单词中的单词

没有准确的输入数据写的就有点虚。

字典树，先建树，再把每个单词分开两部分找，是不是有这样的出现过。

因为输入是按字典序的所以也不用自己再来排序什么的，直接用输入的顺序就行了。

#include
#include
#include
#include
#include
using namespace std;
char s[50000+100][100];
int cnt = 0;
struct node
{
    bool flag;
    char c;
    vectornext;
    node()
    {
        flag = false;
        next.clear();
        c = 0;
    }
}*tree,*add;
bool find_and_insert(node *now,char *str,int x,int len,int ope)
{
    if(x==len)
    {
        if(ope)
            now->flag = true;
        return now->flag;
    }
    for(int i=0;inext.size();i++)
    {
        if(now->next[i]->c!=str[x])
            continue;
        return find_and_insert(now->next[i], str,x+1,len, ope);
    }
    if(ope)
    {
        add = new node;
        add->c = str[x];
        now->next.push_back(add);
        return find_and_insert(add, str,x+1,len, ope);
    }
    return 0;
}
int main()
{
    tree = new node;
    while(scanf("%s",s[cnt])!=EOF)
        find_and_insert(tree,s[cnt],0,(int)strlen(s[cnt]),1),cnt++;
    for(int i=0;i