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



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
Your output should contain all the hat’s words, one per line, in alphabetical order.

[align=left]Sample Input[/align]

a
ahat
hat
hatword
hziee
word

[align=left]Sample Output[/align]

ahat
hatword

[align=left]Author[/align]
戴帽子的

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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
[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;
}