HDU1247：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): 14599    Accepted Submission(s): 5214

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

 

Author

戴帽子的

 

题意：给定一堆单词，若一个单词能由其他两个单词拼接而成，就输出它。

思路：先将所有单词存到字典树里，再遍历每个单词，看能不能有两个组成。

# include <iostream>
# include <cstdio>
# include <cstring>
# define maxn 50000
using namespace std;
int n=0, cnt=0;
char s[maxn+3][20];

struct node
{
    int next[27], tag;
}tri[maxn+100]={0};

void add(char *s)
{
    int len = strlen(s), pos=0;
    for(int i=0; i<len; ++i)
    {
        int t = s[i]-'a';
        if(!tri[pos].next[t])
            tri[pos].next[t] = ++cnt;
        pos = tri[pos].next[t];
    }
    tri[pos].tag = 1;//tag为1表示存在以pos的节点结尾的单词。
}

bool query(char *s)
{
    int len = strlen(s), pos=0;
    for(int i=0; i<=len ;++i)
    {
        int t = s[i]-'a';
        if(tri[pos].next[t])
        {
            pos = tri[pos].next[t];
            if(tri[pos].tag)
            {
                int pos2 = 0, j=-1;
                for(j=i+1; j<len; ++j)
                {
                    int t2 = s[j]-'a';
                    if(!tri[pos2].next[t2]) break;
                    pos2 = tri[pos2].next[t2];
                    if(j == len-1 && tri[pos2].tag)
                        return true;
                }
            }
        }
       else
            return false;
    }
    return false;
}

int main()
{
    while(~scanf("%s",s[++n])) add(s[n]);
    for(int i=1; i<=n; ++i)
        if(query(s[i]))
            printf("%s\n",s[i]);
    return 0;
}