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): 9574    Accepted Submission(s): 3421

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

 

Recommend

题意：推断一个单词能不能有两个单词组成，能够的话就输出。

题解：全部单词建成一颗字典树，单词插入结尾记录单词的个数。查询时直接暴力拆单词推断。

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

#define N 50020

using namespace std;

char s[N][42];

struct Trie {
    int num;
    struct Trie *nxt[26];
    Trie() {
        num=0;
        for(int i=0; i<26; i++) {
            nxt[i]=NULL;
        }
    }
};

void Trie_Inser(Trie *p,char s[]) {
    int i=0;
    Trie *q=p;
    while(s[i]) {
        int nx=s[i]-'a';
        if(q->nxt[nx]==NULL) {
            q->nxt[nx]=new Trie;
        }
        i++;
        q=q->nxt[nx];
    }
    q->num+=1;
}

bool Trie_Serch(Trie *p,char s[],int be,int en) {
    Trie *q=p;
    int i=be;
    while(i<=en) {
        int nx=s[i]-'a';
        if(q->nxt[nx]==NULL)return false;
        q=q->nxt[nx];
        i++;
    }
    if(q->num>0)return true;
    return 0;
}

int main() {
    //freopen("test.in","r",stdin);
    Trie *p=new Trie;
    int id=1;
    while(~scanf("%s",s[id])) {
        Trie_Inser(p,s[id]);
        id++;
    }
    for(int i=1; i<id; i++) {
        int len=strlen(s[i]);
        int be=0;
        if(len<=1)continue;
        for(int en=0; en<len-1; en++) {
            if(Trie_Serch(p,s[i],be,en)&&Trie_Serch(p,s[i],en+1,len-1)) {
                printf("%s\n",s[i]);
                break;
            }
        }
    }
    return 0;
}