hdu–1247–Hat’s Words（一般）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5772    Accepted Submission(s): 2154

Problem DescriptionA 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. 

InputStandard 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. 

OutputYour output should contain all the hat’s words, one per line, in alphabetical order. 

Sample Input

aahathathatwordhzieeword

 

Sample Output

ahathatword    
#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXN=26;const int MAX=50005;char word[MAX][27];struct node{   bool is;   node *next[MAXN];   node()   {      is=false;      memset(next,0,sizeof(next));   }};void Insert(node *rt,char *s){    int i=0;    node *p=rt;    while(s[i])    {         int k=s[i++]-'a';         if(p->next[k]==NULL)            p->next[k]=new node();         p=p->next[k];    }    p->is=true; //該結點是單詞的尾
}bool Search(node *rt,char s[]){int i=0,top=0,stack[1000];node *p=rt;while(s[i]){int k=s[i++]-'a';if(p->next[k]==NULL) return 0;p=p->next[k];if(p->is && s[i])//找到該單詞含有子單詞的分割點 stack[top++]=i;//入棧}while(top)//從可能的分割點去找
{    bool ok=1;i=stack[--top];p=rt;while(s[i]){int k=s[i++]-'a';if(p->next[k]==NULL){ok=false;break;}p=p->next[k];}if(ok && p->is)//找到最後，並且是單詞的return 1;}return 0;}int main(){    int i=0;    node *rt=new node();    while(gets(word[i]))    {        Insert(rt,word[i]);        i++;}for(int j=0;j<i;j++)if(Search(rt,word[j]))   printf("%s\n",word[j]);    return 0;}