【HDU】 1247 Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11664    Accepted Submission(s): 4160

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

 

题解：首先把所有字符串存进Trie里，然后暴力每个字符串.....

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

using namespace std;
struct Trie
{
    Trie *ch[30];
    int val;
};
Trie *root;
char s[50005][30],c1[30],c2[30];

void buildtrie(char *s)
{
    int l=strlen(s),k;
    Trie *p=root,*q;
    for (int i=0;i<l;i++)
    {
        k=s[i]-97;
        if (p->ch[k]==NULL)
        {
            q=(Trie*)malloc(sizeof(Trie));
            for (int j=0;j<26;j++) q->ch[j]=NULL;
            q->val=0;
            p->ch[k]=q;
            p=p->ch[k];
        }
        else p=p->ch[k];
    }
    p->val=1;
}

int intrie(char *s)
{
    int l=strlen(s),k;
    Trie *p=root;
    for (int i=0;i<l;i++)
    {
        k=s[i]-97;
        if (p->ch[k]!=NULL) p=p->ch[k];
        else return 0;
    }
    if (p->val==1) return 1;
}

void solve(int j,int k,int l)
{
    int i=0;
    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));
    for (i=0;i<=k;i++) c1[i]=s[j][i];
    i=0;
    for (int z=k+1;z<l;i++,z++) c2[i]=s[j][z];
}

int main()
{
    int i=0; root=(Trie*)malloc(sizeof(Trie));
    for (i=0;i<26;i++) root->ch[i]=NULL;
    root->val=0;
    i=0;
    while (scanf("%s",s[i])!=EOF)
    {
        buildtrie(s[i]);
        i++;
    }
    for (int j=0;j<i;j++)
    {
        int l=strlen(s[j]),f=0;
        while (f<l-1)
        {
            solve(j,f,strlen(s[j]));
            if (intrie(c1) && intrie(c2))
            {
                printf("%s\n",s[j]);
                break;
            }
            f++;
        }
    }
    return 0;
}