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): 2205    Accepted Submission(s): 804

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：给一些单词（以字典序输入），找出那些可以分成另外的两个单词的单词，以字典序输出；

源代码：（15MS）

#include<iostream>#include<string.h>using namespace std;const int NODEMAX=26;const int MAX=50005;char word[MAX][16];  //这道题16就足够了struct TrieNode{   bool is;   TrieNode *next[NODEMAX];   TrieNode()   {      is=false;      memset(next,0,sizeof(next));   }};void InsertTrie(TrieNode *pRoot,char s[]){    int i;    TrieNode *p=pRoot;    i=0;    while(s[i])    {         int k=s[i]-'a';         if(p->next[k]==NULL)            p->next[k]=new TrieNode();         i++;         p=p->next[k];    }    p->is=true; //该结点是单词的尾}bool Search(TrieNode *pRoot,char s[]){int i,top=0,stack[1000];TrieNode *p=pRoot;i=0;while(s[i]){int k=s[i++]-'a';if(p->next[k]==NULL)return 0;p=p->next[k];if(p->is && s[i]) //找到该单词含有子单词的分隔点stack[top++]=i;//入栈}while(top)//从可能的分割点去找{    bool ok=1;i=stack[--top];p=pRoot;while(s[i]){int k=s[i++]-'a';if(p->next[k]==NULL){ok=false;break;}p=p->next[k];}if(ok && p->is)//找到最后,并且是单词的return 1;}return 0;}int main(){    int i=0;    TrieNode *pRoot=new TrieNode();    while(gets(word[i]))    {        InsertTrie(pRoot,word[i]);        i++;}for(int j=0;j<i;j++)if(Search(pRoot,word[j]))cout<<word[j]<<endl;    return 0;}