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hdu1247Hat’s Words (组合单词，字典树+DFS)

2024-07-14 03:54:18   228人阅读

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

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
typedef struct nn
{
    int flag;
    struct nn *next[26];
}node;
typedef struct strin
{
    int len,indx;
    char str[105];
}String;
int cmp(String s1,String s2)
{
    return s1.len<s2.len;
}
int cmp1(String s1,String s2)
{
    return s1.indx<s2.indx;
}
node *builde()
{
    node *p=new node;
    p->flag=0;
    for(int i=0;i<26;i++)
    p->next[i]=NULL;
    return p;
}
node *root;
void insert(char s[])
{
    node *p=root;
    for(int i=0;s[i]!='\0';i++)
    {
        if(p->next[s[i]-'a']==NULL)
        p->next[s[i]-'a']=builde();
        p=p->next[s[i]-'a'];
    }
    p->flag=1;
}
String s[50005];
int flog;
void dfs(int k,int si,int num)
{
    node *p=root;
    if(num>2)
         return ;
    if(si==s[k].len)
    {
        if(num==2)flog=1; return ;
    }

    for(int j=si;j<s[k].len;j++)
    {
        if(p->next[s[k].str[j]-'a']==NULL)
            return ;
        p=p->next[s[k].str[j]-'a'];
        if(p->flag)
        {
            dfs(k,j+1,num+1); if(flog!=0) return ;
        }
    }
}
int main()
{
    int n=0,m=0;
    while(scanf("%s",s[n].str)>0&&s[n].str[0]!='#')
    {
        s[n].len=strlen(s[n].str); s[n].indx=n; n++;
    }
    sort(s,s+n,cmp);
    root=builde();
    for(int i=0;i<n;i++)
    {
       flog=0; dfs(i,0,0);
       if(flog==1)s[m++]=s[i];
       insert(s[i].str);
    }
    sort(s,s+m,cmp1);
    for(int i=0;i<m;i++)
    printf("%s\n",s[i].str);
}



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